{"id":4956,"date":"2016-05-19T09:49:00","date_gmt":"2016-05-19T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/"},"modified":"2023-03-22T15:19:38","modified_gmt":"2023-03-22T09:49:38","slug":"ncert-solutions-class-10-maths-exercise-5-3","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 5.3"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#NCERT_Solutions_for_Class_10_Maths_Arithmetic_Progressions\" >NCERT Solutions for Class 10 Maths Arithmetic Progressions<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#1_Find_the_sum_of_the_following_APs\" >1. Find the sum of the following AP&#8217;s.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#4_How_many_terms_of_the_AP_9_17_25_%E2%80%A6_must_be_taken_to_give_a_sum_of_636\" >4. How many terms of the AP: 9, 17, 25, &#8230; must be taken to give a sum of 636?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#5_The_first_term_of_an_AP_is_5_the_last_term_is_45_and_the_sum_is_400_Find_the_number_of_terms_and_the_common_difference\" >5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#6_The_first_and_the_last_terms_of_an_AP_are_17_and_350_respectively_If_the_common_difference_is_9_how_many_terms_are_there_and_what_is_their_sum\" >6. The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#7_Find_the_sum_of_first_22_terms_of_an_AP_in_which_d_7_and_22nd_term_is_149\" >7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#8_Find_the_sum_of_first_51_terms_of_an_AP_whose_second_and_third_terms_are_14_and_18_respectively\" >8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#9_If_the_sum_of_first_7_terms_of_an_AP_is_49_and_that_of_17_terms_is_289_find_the_sum_of_first_n_terms\" >9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#10_Show_that_form_an_AP_where_an_is_defined_as_below\" >10. Show that form an AP where an is defined as below:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#11_If_the_sum_of_the_first_n_terms_of_an_AP_is_what_is_the_first_term_that_is_What_is_the_sum_of_first_two_terms_What_is_the_second_term_Similarly_find_the_3rd_the_10th_and_the_nth_terms\" >11. If the sum of the first n terms of an AP is , what is the first term (that is)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#12_Find_the_sum_of_the_first_40_positive_integers_divisible_by_6\" >12. Find the sum of the first 40 positive integers divisible by 6.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#13_Find_the_sum_of_the_first_15_multiples_of_8\" >13. Find the sum of the first 15 multiples of 8.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#14_Find_the_sum_of_the_odd_numbers_between_0_and_50\" >14. Find the sum of the odd numbers between 0 and 50.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#15_A_contract_on_construction_job_specifies_a_penalty_for_delay_of_completion_beyond_a_certain_date_as_follows_Rs_200_for_the_first_day_Rs_250_for_the_second_day_Rs_300_for_the_third_day_etc_the_penalty_for_each_succeeding_day_being_Rs_50_more_than_for_the_preceding_day_How_much_money_the_contractor_has_to_pay_as_penalty_if_he_has_delayed_the_work_by_30_days\" >15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#16_A_sum_of_Rs_700_is_to_be_used_to_give_seven_cash_prizes_to_students_of_a_school_for_their_overall_academic_performance_If_each_prize_is_Rs_20_less_than_its_preceding_term_find_the_value_of_each_of_the_prizes\" >16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 20 less than its preceding term, find the value of each of the prizes.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#17_In_a_school_students_thought_of_planting_trees_in_and_around_the_school_to_reduce_air_pollution_It_was_decided_that_the_number_of_trees_that_each_section_of_each_class_will_plant_will_be_the_same_as_the_class_in_which_they_are_studying_eg_a_section_of_Class_I_will_plant_1_tree_a_section_of_class_II_will_plant_two_trees_and_so_on_till_Class_XII_There_are_three_sections_of_each_class_How_many_trees_will_be_planted_by_the_students\" >17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of Class I will plant 1 tree, a section of class II will plant two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#18_A_spiral_is_made_up_of_successive_semicircles_with_centers_alternatively_at_A_and_B_starting_with_center_at_A_of_radii_05_cm_10_cm_15_cm_20_cm_%E2%80%A6_What_is_the_total_length_of_such_a_spiral_made_up_of_thirteen_consecutive_semicircles\" >18. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, &#8230; What is the total length of such a spiral made up of thirteen consecutive semicircles.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#19_200_logs_are_stacked_in_the_following_manner_20_logs_in_the_bottom_row_19_in_the_next_row_18_in_the_row_next_to_it_and_so_on_In_how_many_rows_are_the_200_logs_placed_and_how_many_logs_are_in_the_top_row\" >19. 200 logs are stacked in the following manner:20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#20_In_a_potato_race_a_bucket_is_placed_at_the_starting_point_which_is_5_meters_from_the_first_potato_and_the_other_potatoes_are_placed_3_meters_apart_in_a_straight_line_There_are_ten_potatoes_in_the_line_A_competitor_starts_from_the_bucket_picks_up_the_nearest_potato_runs_back_with_it_drops_it_in_the_bucket_runs_back_to_pick_up_the_next_potato_runs_to_the_bucket_to_drop_it_in_and_she_continues_in_the_same_way_until_all_the_potatoes_are_in_the_bucket_What_is_the_total_distance_the_competitor_has_to_run\" >20. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#NCERT_Solutions_for_Class_10_Maths_Exercise_53\" >NCERT Solutions for Class 10 Maths Exercise 5.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 5.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Maths\u00a0<\/strong><strong>Arithmetic Progression\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-arithmetic-progressions\/1207\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 5.3\" width=\"163\" height=\"201\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Arithmetic_Progressions\"><\/span>NCERT Solutions for Class 10 Maths Arithmetic Progressions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Find_the_sum_of_the_following_APs\"><\/span><strong>1. Find the sum of the following AP&#8217;s.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) 2, 7, 12&#8230; to 10 terms<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) \u201337, \u201333, \u201329&#8230; to 12 terms<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) 0.6, 1.7, 2.8&#8230; to 100 terms<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) <img decoding=\"async\" style=\"height: 42px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image001.png\" \/> to 11 terms<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> 2, 7, 12&#8230; to 10 terms<\/p>\n<p style=\"text-align: justify;\">Here First term = a = 2, Common difference = d = 7 \u2013 2 = 5 and n = 10<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 302px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image003.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> \u201337, \u201333, \u201329&#8230; to 12 terms<\/p>\n<p style=\"text-align: justify;\">Here First term = a = \u201337, Common difference = d = \u201333 \u2013 (\u201337) = 4<\/p>\n<p style=\"text-align: justify;\">And n = 12<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 368px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> 0.6, 1.7, 2.8&#8230; to 100 terms<\/p>\n<p style=\"text-align: justify;\">Here First term = a = 0.6, Common difference = d = 1.7 \u2013 0.6 = 1.1<\/p>\n<p style=\"text-align: justify;\">And n = 100<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 409px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<\/strong> <img decoding=\"async\" style=\"height: 42px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image001.png\" \/> to 11 terms<\/p>\n<p style=\"text-align: justify;\">Here First tern =a = <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image006.png\" \/> Common difference = d = <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 45px; width: 431px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image008.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: justify;\"><strong>2. Find the sums given below:<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(i) <img decoding=\"async\" style=\"height: 42px; width: 138px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image009.png\" \/> <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) 34 + 32 + 30 + \u2026 + 10<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) \u20135 + (\u20138) + (\u201311) + &#8230; + (\u2013230)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> <img decoding=\"async\" style=\"height: 42px; width: 138px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image009.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here First term = a = 7, Common difference = d = <img decoding=\"async\" style=\"height: 42px; width: 169px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\">And Last term = <em>l <\/em>= 84<\/p>\n<p style=\"text-align: justify;\">We do not know how many terms are there in the given AP.<\/p>\n<p style=\"text-align: justify;\">So, we need to find n first.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">[7 + (<em>n <\/em>\u2212 1) (3.5)] = 84<\/p>\n<p style=\"text-align: justify;\">\u21d2 7 + (3.5) <em>n <\/em>\u2212 3.5 = 84<\/p>\n<p style=\"text-align: justify;\">\u21d2 3.5<em>n <\/em>= 84 + 3.5 \u2013 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 3.5<em>n <\/em>= 80.5<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>n <\/em>= 23<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 23 terms in the given AP.<\/p>\n<p style=\"text-align: justify;\">It means n = 23.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image012.png\" \/>to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 42px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image013.png\" \/> <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em><img decoding=\"async\" style=\"height: 42px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image014.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> 34 + 32 + 30 + &#8230; + 10<\/p>\n<p style=\"text-align: justify;\">Here First term = a = 34, Common difference = d = 32 \u2013 34 = \u20132<\/p>\n<p style=\"text-align: justify;\">And Last term = <em>l <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">We do not know how many terms are there in the given AP.<\/p>\n<p style=\"text-align: justify;\">So, we need to find n first.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">[34 + (<em>n <\/em>\u2212 1) (\u22122)] = 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 34 \u2013 2<em>n <\/em>+ 2 = 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22122<em>n <\/em>= \u221226\u21d2 <em>n <\/em>= 13<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 13 terms in the given AP.<\/p>\n<p style=\"text-align: justify;\">It means n = 13.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image012.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 42px; width: 211px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image015.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> \u22125 + (\u22128) + (\u221211) + &#8230; + (\u2212230)<\/p>\n<p style=\"text-align: justify;\">Here First term = a = \u20135, Common difference = d = \u20138 \u2013 (\u20135) = \u20138 + 5 = \u20133<\/p>\n<p style=\"text-align: justify;\">And Last term = <em>l <\/em>= \u2212230<\/p>\n<p style=\"text-align: justify;\">We do not know how many terms are there in the given AP.<\/p>\n<p style=\"text-align: justify;\">So, we need to find n first.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">[\u22125 + (<em>n <\/em>\u2212 1) (\u22123)] = \u2212230<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22125 \u2212 3<em>n <\/em>+ 3 = \u2212230<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22123<em>n <\/em>= \u2212228 \u21d2 <em>n <\/em>= 76<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 76 terms in the given AP.<\/p>\n<p style=\"text-align: justify;\">It means n = 76.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image012.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 42px; width: 269px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image016.png\" \/><\/em><\/p>\n<hr \/>\n<p style=\"text-align: justify;\"><strong>3. In an AP<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(i) given <em>a <\/em>= 5, <em>d <\/em>= 3, <img decoding=\"async\" style=\"height: 24px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image017.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image018.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) given <em>a <\/em>= 7, <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image019.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image020.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) given <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image021.png\" \/>, <em>d <\/em>= 3, find<img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image022.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) given <img decoding=\"async\" style=\"height: 24px; width: 124px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image023.png\" \/>, find<\/strong><img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image024.png\" \/><strong>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(v) given <em>d <\/em>= 5, <img decoding=\"async\" style=\"height: 24px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image025.png\" \/>,find<img decoding=\"async\" style=\"height: 24px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image026.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(vi) given <em>a <\/em>= 2, <em>d <\/em>= 8, <img decoding=\"async\" style=\"height: 24px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image027.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image028.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(vii) given <em>a <\/em>= 8, <img decoding=\"async\" style=\"height: 24px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image029.png\" \/>, find n and d.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(viii) given <img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image030.png\" \/>, <em>d <\/em>= 2, <img decoding=\"async\" style=\"height: 24px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image031.png\" \/>, find n and a.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ix) given <em>a <\/em>= 3, <em>n <\/em>= 8, <em>S <\/em>= 192, find d.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(x) given <em>l <\/em>= 28, <em>S <\/em>= 144, and there are total of 9 terms. Find a.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> Given <em>a <\/em>= 5, <em>d <\/em>= 3, <img decoding=\"async\" style=\"height: 24px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image032.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image033.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image034.png\" \/>= 5 + (<em>n <\/em>\u2212 1) (3)<\/p>\n<p style=\"text-align: justify;\">\u21d2 50 = 5 + 3<em>n <\/em>\u2212 3<\/p>\n<p style=\"text-align: justify;\">\u21d2 48 = 3<em>n<\/em>\u21d2 <em>n <\/em>= 16<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 316px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image035.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 16 and <img decoding=\"async\" style=\"height: 24px; width: 65px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image036.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> Given <em>a <\/em>= 7, <img decoding=\"async\" style=\"height: 24px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image037.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image038.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image039.png\" \/>= 7 + (13 \u2212 1) (<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 35 = 7 + 12<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 28 = 12<em>d<\/em>\u21d2 <em>d <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image040.png\" \/><\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 45px; width: 347px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image041.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>d <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image040.png\" \/> and <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image042.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> Given <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image043.png\" \/>, <em>d <\/em>= 3, find<img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image044.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image045.png\" \/>= <em>a <\/em>+ (12 \u2212 1) 3<\/p>\n<p style=\"text-align: justify;\">\u21d2 37 = <em>a <\/em>+ 33 \u21d2 <em>a <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 302px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image046.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>a <\/em>= 4 and <em>S<\/em><sub>12<\/sub> = 246<\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<\/strong> Given <img decoding=\"async\" style=\"height: 24px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image047.png\" \/>, find<img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image048.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image049.png\" \/>= <em>a <\/em>+ (3 \u2212 1) (<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 15 = <em>a <\/em>+ 2<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= 15 \u2212 2<em>d\u2026<\/em> (1)<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 152px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image050.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 125 = 5 (2<em>a <\/em>+ 9<em>d<\/em>) = 10<em>a <\/em>+ 45<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting (1) in the above equation,<\/p>\n<p style=\"text-align: justify;\">125 = 5 [2 (15 \u2212 2<em>d<\/em>) + 9<em>d<\/em>] = 5 (30 \u2212 4<em>d <\/em>+ 9<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 125 = 150 + 25<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 125 \u2013 150 = 25<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 \u221225 = 25<em>d<\/em>\u21d2 <em>d <\/em>= \u22121<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image051.png\" \/>= <em>a <\/em>+ (10 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> and equation (1) in the above equation,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image051.png\" \/>= 15 \u2212 2<em>d <\/em>+ 9<em>d <\/em>= 15 + 7<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">= 15 + 7 (\u22121) = 15 \u2013 7 = 8<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>d <\/em>= \u22121 and <img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image051.png\" \/> = 8<\/p>\n<p style=\"text-align: justify;\"><strong>(v)<\/strong> Given <em>d <\/em>= 5,<img decoding=\"async\" style=\"height: 24px; width: 149px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image052.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image053.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 104px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image054.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 150 = 18<em>a <\/em>+ 360<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u2212210 = 18<em>a <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image055.png\" \/><\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image056.png\" \/>= <img decoding=\"async\" style=\"height: 42px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image055.png\" \/> + (9 \u2212 1) (5)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 181px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image057.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>a <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image055.png\" \/> and <em>a<\/em><sub>9<\/sub> = <img decoding=\"async\" style=\"height: 42px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image058.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(vi)<\/strong> Given <em>a <\/em>= 2, <em>d <\/em>= 8,<img decoding=\"async\" style=\"height: 24px; width: 152px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image059.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image060.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 116px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image061.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image062.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image063.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image064.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 175px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image065.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>n <\/em>(<em>n <\/em>\u2212 5) + 9 (<em>n <\/em>\u2212 5) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 (<em>n <\/em>\u2212 5) (2<em>n <\/em>+ 9) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>n <\/em>= 5,\u22129\/2<\/p>\n<p style=\"text-align: justify;\">We discard negative value of n because here n cannot be in negative or fraction.<\/p>\n<p style=\"text-align: justify;\">The value of n must be a positive integer.<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>,to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a<\/em><sub>5<\/sub> = 2 + (5 \u2212 1) (8) = 2 + 32 = 34<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 5 and <img decoding=\"async\" style=\"height: 24px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image066.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(vii)<\/strong> Given <em>a <\/em>= 8, <img decoding=\"async\" style=\"height: 24px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image067.png\" \/>, find n and d.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">62 = 8 + (<em>n <\/em>\u2212 1) (<em>d<\/em>) = 8 + <em>nd <\/em>\u2013 <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 62 = 8 + <em>nd <\/em>\u2212 <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>nd <\/em>\u2013 <em>d <\/em>= 54<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>nd <\/em>= 54 + <em>d\u2026 <\/em>(1)<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 253px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image068.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting equation (1) in the above equation,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 214px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image069.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image070.png\" \/>\u21d2 <em>n <\/em>= 6<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>n<\/em> in equation (1),<\/p>\n<p style=\"text-align: justify;\">6<em>d <\/em>= 54 + <em>d <\/em>\u21d2 <em>d <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image071.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 6 and <em>d <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image071.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(viii)<\/strong> Given <img decoding=\"async\" style=\"height: 24px; width: 164px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image072.png\" \/>, find n and a.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">4 = <em>a <\/em>+ (<em>n <\/em>\u2212 1) (2) = <em>a <\/em>+ 2<em>n <\/em>\u2212 2<\/p>\n<p style=\"text-align: justify;\">\u21d2 4 = <em>a <\/em>+ 2<em>n <\/em>\u2013 2<\/p>\n<p style=\"text-align: justify;\">\u21d2 6 = <em>a <\/em>+ 2<em>n <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= 6 \u2212 2<em>n\u2026<\/em> (1)<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 256px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image073.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image074.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting equation (1) in the above equation, we get<\/p>\n<p style=\"text-align: justify;\">\u221228 = <em>n <\/em>[2 (6 \u2212 2<em>n<\/em>) + 2<em>n <\/em>\u2013 2]\n<p style=\"text-align: justify;\">\u21d2 \u221228 = <em>n <\/em>(12 \u2212 4<em>n <\/em>+ 2<em>n <\/em>\u2212 2)<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u221228 = <em>n <\/em>(10 \u2212 2<em>n<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image075.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image076.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 160px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image077.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>n <\/em>(<em>n <\/em>\u2212 7) + 2 (<em>n <\/em>\u2212 7) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 (<em>n <\/em>+ 2) (<em>n <\/em>\u2212 7) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>n <\/em>= \u22122, 7<\/p>\n<p style=\"text-align: justify;\">Here, we cannot have negative value of <em>n<\/em>.<\/p>\n<p style=\"text-align: justify;\">Therefore, we discard negative value of <em>n<\/em> which means <em>n <\/em>= 7.<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>n<\/em> in equation (1), we get<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>= 6 \u2212 2<em>n <\/em>= 6 \u2013 2 (7) = 6 \u2013 14 = \u22128<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 7 and <em>a <\/em>= \u22128<\/p>\n<p style=\"text-align: justify;\"><strong>(ix)<\/strong>Given <em>a <\/em>= 3, <em>n <\/em>= 8, <em>S <\/em>= 192, find d.<\/p>\n<p style=\"text-align: justify;\">Using formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\">192 = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image078.png\" \/>[6 + (8 \u2212 1) <em>d<\/em>] = 4 (6 + 7<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 192 = 24 + 28<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 168 = 28<em>d <\/em>\u21d2 <em>d <\/em>= 6<\/p>\n<p style=\"text-align: justify;\"><strong>(x)<\/strong> Given <em>l <\/em>= 28, <em>S <\/em>= 144, and there are total of 9 terms. Find a.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image079.png\" \/>, to find sum of n terms, we get<\/p>\n<p style=\"text-align: justify;\">144 = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image080.png\" \/> [<em>a <\/em>+ 28]\n<p style=\"text-align: justify;\">\u21d2 288 = 9 [<em>a <\/em>+ 28]\n<p style=\"text-align: justify;\">\u21d2 32 = <em>a <\/em>+ 28\u21d2 <em>a <\/em>= 4<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_How_many_terms_of_the_AP_9_17_25_%E2%80%A6_must_be_taken_to_give_a_sum_of_636\"><\/span><strong>4. How many terms of the AP: 9, 17, 25, &#8230; must be taken to give a sum of 636?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>First term = a = 9, Common difference = d = 17 \u2013 9 = 8, <em>S<sub>n<\/sub><\/em> = 636<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\">636 = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image081.png\" \/>[18 + (<em>n <\/em>\u2212 1) (8)]\n<p style=\"text-align: justify;\">\u21d2 1272 = <em>n <\/em>(18 + 8<em>n <\/em>\u2212 8)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image082.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image083.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image084.png\" \/><\/p>\n<p style=\"text-align: justify;\">Comparing equation <img decoding=\"async\" style=\"height: 22px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image084.png\" \/>with general form <img decoding=\"async\" style=\"height: 22px; width: 108px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image085.png\" \/>, we get<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>= 4, <em>b <\/em>= 5 and <em>c <\/em>= \u2212636<\/p>\n<p style=\"text-align: justify;\">Applying quadratic formula, <img decoding=\"async\" style=\"height: 31px; width: 103px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image086.png\" \/>and putting values of a, b and c, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 32px; width: 143px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image087.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 30px; width: 115px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image088.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 87px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image089.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image090.png\" \/><\/p>\n<p style=\"text-align: justify;\">We discard negative value of <em>n<\/em> here because n cannot be in negative, n can only be a positive integer.<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>n <\/em>= 12<\/p>\n<p style=\"text-align: justify;\">Therefore, 12 terms of the given sequence make sum equal to 636.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_The_first_term_of_an_AP_is_5_the_last_term_is_45_and_the_sum_is_400_Find_the_number_of_terms_and_the_common_difference\"><\/span><strong>5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>First term = a = 5, Last term = <em>l <\/em>= 45, <img decoding=\"async\" style=\"height: 24px; width: 65px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image091.png\" \/><\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image079.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image092.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image093.png\" \/>\u21d2 n = 16<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP and putting value of n, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image094.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 400 = 8 (10 + 15<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 400 = 80 + 120<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 320 = 120<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>d <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image095.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_The_first_and_the_last_terms_of_an_AP_are_17_and_350_respectively_If_the_common_difference_is_9_how_many_terms_are_there_and_what_is_their_sum\"><\/span><strong>6. The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>First term = <em>a <\/em>= 17, Last term = <em>l <\/em>= 350 and Common difference = <em>d <\/em>= 9<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find nth term of arithmetic progression, we get<\/p>\n<p style=\"text-align: justify;\">350 = 17 + (<em>n <\/em>\u2212 1) (9)<\/p>\n<p style=\"text-align: justify;\">\u21d2 350 = 17 + 9<em>n <\/em>\u2212 9<\/p>\n<p style=\"text-align: justify;\">\u21d2 342 = 9<em>n <\/em>\u21d2 <em>n <\/em>= 38<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP and putting value of n, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image096.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image097.png\" \/>= 19 (34 + 333) = 6973<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 38 terms and their sum is equal to 6973.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_Find_the_sum_of_first_22_terms_of_an_AP_in_which_d_7_and_22nd_term_is_149\"><\/span><strong>7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that 22nd term is equal to 149 <img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image098.png\" \/> <img decoding=\"async\" style=\"height: 24px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image099.png\" \/><\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find nth term of arithmetic progression, we get<\/p>\n<p style=\"text-align: justify;\">149 = <em>a <\/em>+ (22 \u2212 1) (7)<\/p>\n<p style=\"text-align: justify;\">\u21d2 149 = <em>a <\/em>+ 147\u21d2 <em>a <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP and putting value of a, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image100.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image101.png\" \/>= 11 (4 + 147)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image101.png\" \/>= 1661<\/p>\n<p style=\"text-align: justify;\">Therefore, sum of first 22 terms of AP is equal to 1661.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_Find_the_sum_of_first_51_terms_of_an_AP_whose_second_and_third_terms_are_14_and_18_respectively\"><\/span><strong>8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that second and third term of AP are 14 and 18 respectively.<\/p>\n<p style=\"text-align: justify;\">Using formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image011.png\" \/>, to find nth term of arithmetic progression, we get<\/p>\n<p style=\"text-align: justify;\">14 = <em>a <\/em>+ (2 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 14 = <em>a <\/em>+ <em>d \u2026 <\/em>(1)<\/p>\n<p style=\"text-align: justify;\">And, 18 = <em>a <\/em>+ (3 \u2212 1) <em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 18 = <em>a <\/em>+ 2<em>d \u2026 <\/em>(2)<\/p>\n<p style=\"text-align: justify;\">These are equations consisting of two variables.<\/p>\n<p style=\"text-align: justify;\">Using equation (1), we get, <em>a <\/em>= 14 \u2212 <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting value of <em>a<\/em> in equation (2), we get<\/p>\n<p style=\"text-align: justify;\">18 = 14 \u2013 <em>d <\/em>+ 2<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>d <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">Therefore, common difference <em>d <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in equation (1), we get<\/p>\n<p style=\"text-align: justify;\">18 = <em>a <\/em>+ 2 (4)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 434px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image102.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, sum of first 51 terms of an AP is equal to 5610.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_If_the_sum_of_first_7_terms_of_an_AP_is_49_and_that_of_17_terms_is_289_find_the_sum_of_first_n_terms\"><\/span><strong>9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 135px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image103.png\" \/>49<\/p>\n<p style=\"text-align: justify;\">\u21d2 98 = 7 (2<em>a <\/em>+ 6<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 7 = <em>a <\/em>+ 3<em>d <\/em>\u21d2 <em>a <\/em>= 7 \u2212 3<em>d \u2026 <\/em>(1)<\/p>\n<p style=\"text-align: justify;\">And, <img decoding=\"async\" style=\"height: 42px; width: 156px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image104.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 578 = 17 (2<em>a <\/em>+ 16<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 34 = 2<em>a <\/em>+ 16<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 17 = <em>a <\/em>+ 8<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting equation (1) in the above equation, we get<\/p>\n<p style=\"text-align: justify;\">17 = 7 \u2212 3<em>d <\/em>+ 8<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 10 = 5<em>d <\/em>\u21d2 <em>d <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in equation (1), we get<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>= 7 \u2212 3<em>d <\/em>= 7 \u2013 3 (2) = 7 \u2013 6 = 1<\/p>\n<p style=\"text-align: justify;\">Again applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image105.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image106.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image107.png\" \/>\u21d2 <em>S<sub>n<\/sub> <\/em>= <em>n<\/em><sup>2<\/sup><\/p>\n<p style=\"text-align: justify;\">Therefore, sum of n terms of AP is equal to <em>n<\/em><sup>2<\/sup>.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"10_Show_that_form_an_AP_where_an_is_defined_as_below\"><\/span><strong>10. Show that <img decoding=\"async\" style=\"height: 24px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image108.png\" \/>form an AP where a<sub>n<\/sub> is defined as below:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) <img decoding=\"async\" style=\"height: 24px; width: 91px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image109.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) <img decoding=\"async\" style=\"height: 24px; width: 91px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image110.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Also find the sum of the first 15 terms in each case.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong>We need to show that <img decoding=\"async\" style=\"height: 24px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image111.png\" \/> form an AP where <img decoding=\"async\" style=\"height: 24px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image112.png\" \/><\/p>\n<p style=\"text-align: justify;\">Let us calculate values of <img decoding=\"async\" style=\"height: 24px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image113.png\" \/> using <img decoding=\"async\" style=\"height: 24px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image112.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image114.png\" \/>= <img decoding=\"async\" style=\"height: 24px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image115.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 3 + 4 (2) = 3 + 8 = 11<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image116.png\" \/>= <img decoding=\"async\" style=\"height: 24px; width: 104px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image117.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 3 + 4 (4) = 3 + 16 = 19<\/p>\n<p style=\"text-align: justify;\">So, the sequence is of the form 7, 11, 15, 19 &#8230;<\/p>\n<p style=\"text-align: justify;\">Let us check difference between consecutive terms of this sequence.<\/p>\n<p style=\"text-align: justify;\">11 \u2013 7 = 4, 15 \u2013 11 = 4, 19 \u2013 15 = 4<\/p>\n<p style=\"text-align: justify;\">Therefore, the difference between consecutive terms is constant which means terms <em>a<\/em><sub>1<\/sub>, <em>a<\/em><sub>2<\/sub> &#8230; <em>a<sub>n<\/sub><\/em> form an AP.<\/p>\n<p style=\"text-align: justify;\">We have sequence 7, 11, 15, 19 &#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 7 and Common difference = d = 4<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 394px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image118.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, sum of first 15 terms of AP is equal to 525.<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong>We need to show that <img decoding=\"async\" style=\"height: 24px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image111.png\" \/> form an AP where <img decoding=\"async\" style=\"height: 24px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image119.png\" \/><\/p>\n<p style=\"text-align: justify;\">Let us calculate values of <img decoding=\"async\" style=\"height: 24px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image113.png\" \/> using <img decoding=\"async\" style=\"height: 24px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image119.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image120.png\" \/>= <img decoding=\"async\" style=\"height: 24px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image121.png\" \/>= 9 \u2013 5 (2) = 9 \u2013 10 = \u22121<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image122.png\" \/>= <img decoding=\"async\" style=\"height: 24px; width: 112px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image123.png\" \/>= 9 \u2013 5 (4) = 9 \u2013 20 = \u221211<\/p>\n<p style=\"text-align: justify;\">So, the sequence is of the form 4, \u22121, \u22126, \u221211 &#8230;<\/p>\n<p style=\"text-align: justify;\">Let us check difference between consecutive terms of this sequence.<\/p>\n<p style=\"text-align: justify;\">\u20131 \u2013 (4) = \u20135,\u20136 \u2013 (\u20131)<\/p>\n<p style=\"text-align: justify;\">= \u20136 + 1 = \u20135,\u201311 \u2013 (\u20136)<\/p>\n<p style=\"text-align: justify;\">= \u201311 + 6 = \u20135<\/p>\n<p style=\"text-align: justify;\">Therefore, the difference between consecutive terms is constant which means terms <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image124.png\" \/> form an AP.<\/p>\n<p style=\"text-align: justify;\">We have sequence 4, \u22121, \u22126, \u221211 &#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 4 and Common difference = d = \u20135<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/> to find sum of n terms of AP , we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 449px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image125.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, sum of first 15 terms of AP is equal to \u2013465.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"11_If_the_sum_of_the_first_n_terms_of_an_AP_is_what_is_the_first_term_that_is_What_is_the_sum_of_first_two_terms_What_is_the_second_term_Similarly_find_the_3rd_the_10th_and_the_nth_terms\"><\/span><strong>11. If the sum of the first n terms of an AP is <img decoding=\"async\" style=\"height: 29px; width: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image126.png\" \/>, what is the first term (that is<img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image127.png\" \/>)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that the sum of n terms of an AP is equal to <img decoding=\"async\" style=\"height: 29px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image128.png\" \/><\/p>\n<p style=\"text-align: justify;\">It means <img decoding=\"async\" style=\"height: 25px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image129.png\" \/><\/p>\n<p style=\"text-align: justify;\">Let us calculate <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image130.png\" \/> using <img decoding=\"async\" style=\"height: 25px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image129.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 29px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image131.png\" \/>= 4 \u2013 1 = 3<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 29px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image132.png\" \/>= 8 \u2013 4 = 4<\/p>\n<p style=\"text-align: justify;\">First term = a = <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image133.png\" \/>= 3 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Let us find common difference now.<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<p style=\"text-align: justify;\">We can write any AP in the form of general terms like a , a + d, a + 2d &#8230;<\/p>\n<p style=\"text-align: justify;\">We have calculated that sum of first two terms is equal to 4 i.e. <img decoding=\"async\" style=\"height: 24px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image134.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, we can say that a + (a + d) = 4<\/p>\n<p style=\"text-align: justify;\">Putting value of a from equation (1), we get<\/p>\n<p style=\"text-align: justify;\">2a + d = 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 2 (3) + d = 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 6 + d = 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 d = \u20132<\/p>\n<p style=\"text-align: justify;\">Using formula a<sub>n<\/sub> = a + (n \u2013 1) d, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">Second term of AP = <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image135.png\" \/> = a + (2 \u2013 1) d = 3 + (2 \u2013 1) (\u20132) = 3 \u2013 2 = 1<\/p>\n<p style=\"text-align: justify;\">Third term of AP = <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image049.png\" \/> = a + (3 \u2013 1) d = 3 + (3 \u2013 1) (\u20132) = 3 \u2013 4 = \u20131<\/p>\n<p style=\"text-align: justify;\">Tenth term of AP = <img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image051.png\" \/> = a + (10 \u2013 1) d = 3 + (10 \u2013 1) (\u20132) = 3 \u2013 18 = \u201315<\/p>\n<p style=\"text-align: justify;\">n<sup>th<\/sup> term of AP =<img decoding=\"async\" style=\"height: 24px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image136.png\" \/> = a + (n \u2013 1) d = 3 + (n \u2013 1) (\u20132) = 3 \u2013 2n + 2 = 5 \u2013 2n<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"12_Find_the_sum_of_the_first_40_positive_integers_divisible_by_6\"><\/span><strong>12. Find the sum of the first 40 positive integers divisible by 6.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The first 40 positive integers divisible by 6 are 6, 12, 18, 24 &#8230; 40 terms.<\/p>\n<p style=\"text-align: justify;\">Therefore, we want to find sum of 40 terms of sequence of the form:<\/p>\n<p style=\"text-align: justify;\">6, 12, 18, 24 &#8230; 40 terms<\/p>\n<p style=\"text-align: justify;\">Here, first term = a = 6 and Common difference = d = 12 \u2013 6 = 6, n = 40<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 152px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image137.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 26px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image138.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 20 (12 + 234)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 127px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image139.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"13_Find_the_sum_of_the_first_15_multiples_of_8\"><\/span><strong>13. Find the sum of the first 15 multiples of 8.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The first 15 multiples of 8 are 8,16, 24, 32 &#8230; 15 times<\/p>\n<p style=\"text-align: justify;\">First term = a = 8 and Common difference = d = 16 \u2013 8 = 8, n = 15<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 516px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image140.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"14_Find_the_sum_of_the_odd_numbers_between_0_and_50\"><\/span><strong>14. Find the sum of the odd numbers between 0 and 50.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The odd numbers between 0 and 50 are 1, 3, 5, 7 &#8230; 49<\/p>\n<p style=\"text-align: justify;\">It is an arithmetic progression because the difference between consecutive terms is constant.<\/p>\n<p style=\"text-align: justify;\">First term = a = 1, Common difference = 3 \u2013 1 = 2, Last term = <em>l<\/em> = 49<\/p>\n<p style=\"text-align: justify;\">We do not know how many odd numbers are present between 0 and 50.<\/p>\n<p style=\"text-align: justify;\">Therefore, we need to find n first.<\/p>\n<p style=\"text-align: justify;\">Using formula a<sub>n<\/sub> = a + (n \u2212 1) d, to find nth term of arithmetic progression, we get<\/p>\n<p style=\"text-align: justify;\">49 = 1 + (n \u2212 1) 2<\/p>\n<p style=\"text-align: justify;\">\u21d2 49 = 1 + 2n \u2212 2<\/p>\n<p style=\"text-align: justify;\">\u21d2 50 = 2n \u21d2 n = 25<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image141.png\" \/> to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 272px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image142.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"15_A_contract_on_construction_job_specifies_a_penalty_for_delay_of_completion_beyond_a_certain_date_as_follows_Rs_200_for_the_first_day_Rs_250_for_the_second_day_Rs_300_for_the_third_day_etc_the_penalty_for_each_succeeding_day_being_Rs_50_more_than_for_the_preceding_day_How_much_money_the_contractor_has_to_pay_as_penalty_if_he_has_delayed_the_work_by_30_days\"><\/span><strong>15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Penalty for first day = Rs 200, Penalty for second day = Rs 250<\/p>\n<p style=\"text-align: justify;\">Penalty for third day = Rs 300<\/p>\n<p style=\"text-align: justify;\">It is given that penalty for each succeeding day is Rs 50 more than the preceding day.<\/p>\n<p style=\"text-align: justify;\">It makes it an arithmetic progression because the difference between consecutive terms is constant.<\/p>\n<p style=\"text-align: justify;\">We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.<\/p>\n<p style=\"text-align: justify;\">So, we have an AP of the form200, 250, 300, 350 &#8230; 30 terms<\/p>\n<p style=\"text-align: justify;\">First term = a = 200, Common difference = d = 50, n = 30<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 166px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image143.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 26px; width: 173px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image144.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image145.png\" \/>= 15 (400 + 1450) = 27750<\/p>\n<p style=\"text-align: justify;\">Therefore, penalty for 30 days is Rs. 27750.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"16_A_sum_of_Rs_700_is_to_be_used_to_give_seven_cash_prizes_to_students_of_a_school_for_their_overall_academic_performance_If_each_prize_is_Rs_20_less_than_its_preceding_term_find_the_value_of_each_of_the_prizes\"><\/span><strong>16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 20 less than its preceding term, find the value of each of the prizes.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that sum of seven cash prizes is equal to Rs 700.<\/p>\n<p style=\"text-align: justify;\">And, each prize is R.s 20 less than its preceding term.<\/p>\n<p style=\"text-align: justify;\">Let value of first prize = Rs. a<\/p>\n<p style=\"text-align: justify;\">Let value of second prize =Rs (a \u2212 20)<\/p>\n<p style=\"text-align: justify;\">Let value of third prize = Rs (a \u2212 40)<\/p>\n<p style=\"text-align: justify;\">So, we have sequence of the form:<\/p>\n<p style=\"text-align: justify;\">a, a \u2212 20, a \u2212 40, a \u2013 60 &#8230;<\/p>\n<p style=\"text-align: justify;\">It is an arithmetic progression because the difference between consecutive terms is constant.<\/p>\n<p style=\"text-align: justify;\">First term = a, Common difference = d = (a \u2013 20) \u2013 a = \u201320<\/p>\n<p style=\"text-align: justify;\">n = 7 (Because there are total of seven prizes)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 84px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image146.png\" \/>{given}<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image147.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image148.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 200 = 2a \u2013 120<\/p>\n<p style=\"text-align: justify;\">\u21d2 320 = 2a \u21d2 a = 160<\/p>\n<p style=\"text-align: justify;\">Therefore, value of first prize = Rs 160<\/p>\n<p style=\"text-align: justify;\">Value of second prize = 160 \u2013 20 = Rs 140<\/p>\n<p style=\"text-align: justify;\">Value of third prize = 140 \u2013 20 = Rs 120<\/p>\n<p style=\"text-align: justify;\">Value of fourth prize = 120 \u2013 20 = Rs 100<\/p>\n<p style=\"text-align: justify;\">Value of fifth prize = 100 \u2013 20 = Rs 80<\/p>\n<p style=\"text-align: justify;\">Value of sixth prize = 80 \u2013 20 = Rs 60<\/p>\n<p style=\"text-align: justify;\">Value of seventh prize = 60 \u2013 20 = Rs 40<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"17_In_a_school_students_thought_of_planting_trees_in_and_around_the_school_to_reduce_air_pollution_It_was_decided_that_the_number_of_trees_that_each_section_of_each_class_will_plant_will_be_the_same_as_the_class_in_which_they_are_studying_eg_a_section_of_Class_I_will_plant_1_tree_a_section_of_class_II_will_plant_two_trees_and_so_on_till_Class_XII_There_are_three_sections_of_each_class_How_many_trees_will_be_planted_by_the_students\"><\/span><strong>17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of Class I will plant 1 tree, a section of class II will plant two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>There are three sections of each class and it is given that the number of trees planted by any class is equal to class number.<\/p>\n<p style=\"text-align: justify;\">The number of trees planted by class I = number of sections <img decoding=\"async\" style=\"height: 19px; width: 118px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image149.png\" \/><\/p>\n<p style=\"text-align: justify;\">The number of trees planted by class II = number of sections <img decoding=\"async\" style=\"height: 19px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image150.png\" \/><\/p>\n<p style=\"text-align: justify;\">The number of trees planted by class III = number of sections <img decoding=\"async\" style=\"height: 19px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image151.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, we have sequence of the form 3, 6, 9 &#8230; 12 terms<\/p>\n<p style=\"text-align: justify;\">To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 &#8230; 12 terms.<\/p>\n<p style=\"text-align: justify;\">First term = a = 3, Common difference = d= 6 \u2013 3 = 3 and n = 12<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP , we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 304px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image152.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"18_A_spiral_is_made_up_of_successive_semicircles_with_centers_alternatively_at_A_and_B_starting_with_center_at_A_of_radii_05_cm_10_cm_15_cm_20_cm_%E2%80%A6_What_is_the_total_length_of_such_a_spiral_made_up_of_thirteen_consecutive_semicircles\"><\/span><strong>18. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, &#8230; What is the total length of such a spiral made up of thirteen consecutive semicircles.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 182px; width: 240px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image153.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Length of semi\u2013circle = <img decoding=\"async\" style=\"height: 42px; width: 231px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image154.png\" \/><\/p>\n<p style=\"text-align: justify;\">Length of semi-circle of radii 0.5 cm = <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(0.5) cm<\/p>\n<p style=\"text-align: justify;\">Length of semi-circle of radii 1.0 cm = <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.0) cm<\/p>\n<p style=\"text-align: justify;\">Length of semi-circle of radii 1.5 cm = <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.5) cm<\/p>\n<p style=\"text-align: justify;\">Therefore, we have sequence of the form:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(0.5), <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.0), <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.5) &#8230; 13 terms {There are total of thirteen semi\u2013circles}.<\/p>\n<p style=\"text-align: justify;\">To find total length of the spiral, we need to find sum of the sequence <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(0.5), <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.0), <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.5) &#8230; 13 terms<\/p>\n<p style=\"text-align: justify;\">Total length of spiral = <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(0.5) + <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.0) + <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/>(1.5) &#8230; 13 terms<\/p>\n<p style=\"text-align: justify;\">\u21d2 Total length of spiral =<img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/> (0.5 + 1.0 + 1.5) &#8230; 13 terms \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Sequence 0.5, 1.0, 1.5 &#8230; 13 terms is an arithmetic progression.<\/p>\n<p style=\"text-align: justify;\">Let us find the sum of this sequence.<\/p>\n<p style=\"text-align: justify;\">First term = a = 0.5, Common difference = 1.0 \u2013 0.5 = 0.5 and n = 13<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 322px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image156.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, 0.5 + 1.0 + 1.5 + 2.0 &#8230; 13 terms = 45.5<\/p>\n<p style=\"text-align: justify;\">Putting this in equation (1), we get<\/p>\n<p style=\"text-align: justify;\">Total length of spiral=<img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/> (0.5 + 1.5 + 2.0 + &#8230; 13 terms) = <img decoding=\"async\" style=\"height: 15px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image155.png\" \/> (45.5) = 143 cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"19_200_logs_are_stacked_in_the_following_manner_20_logs_in_the_bottom_row_19_in_the_next_row_18_in_the_row_next_to_it_and_so_on_In_how_many_rows_are_the_200_logs_placed_and_how_many_logs_are_in_the_top_row\"><\/span><strong>19. 200 logs are stacked in the following manner:20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 104px; width: 402px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image157.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The number of logs in the bottom row = 20<\/p>\n<p style=\"text-align: justify;\">The number of logs in the next row = 19<\/p>\n<p style=\"text-align: justify;\">The number of logs in the next to next row = 18<\/p>\n<p style=\"text-align: justify;\">Therefore, we have sequence of the form 20, 19, 18 &#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 20, Common difference = d = 19 \u2013 20 = \u20131<\/p>\n<p style=\"text-align: justify;\">We need to find that how many rows make total of 200 logs.<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 160px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image158.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 400 = n (40 \u2013 n + 1)<\/p>\n<p style=\"text-align: justify;\">\u21d2 400 = <img decoding=\"async\" style=\"height: 22px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image159.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image160.png\" \/><\/p>\n<p style=\"text-align: justify;\">It is a quadratic equation, we can factorize to solve the equation.<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 22px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image161.png\" \/>\u2212 25n \u2212 16n + 400 = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 n (n \u2212 25) \u2013 16 (n \u2212 25) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 (n \u2212 25) (n \u2212 16)<\/p>\n<p style=\"text-align: justify;\">\u21d2 n = 25, 16<\/p>\n<p style=\"text-align: justify;\">We discard n = 25 because we cannot have more than 20 rows in the sequence. The sequence is of the form: 20, 19, 18 &#8230;<\/p>\n<p style=\"text-align: justify;\">At most, we can have 20 or less number of rows.<\/p>\n<p style=\"text-align: justify;\">Therefore, n = 16 which means 16 rows make total number of logs equal to 200.<\/p>\n<p style=\"text-align: justify;\">We also need to find number of logs in the 16th row.<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image141.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\">200 = 8 (20 + <em>l<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 200 = 160 + 8<em>I<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 40 = 8<em>l <\/em>\u21d2 <em>l <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 5 logs in the top most row and there are total of 16 rows.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"20_In_a_potato_race_a_bucket_is_placed_at_the_starting_point_which_is_5_meters_from_the_first_potato_and_the_other_potatoes_are_placed_3_meters_apart_in_a_straight_line_There_are_ten_potatoes_in_the_line_A_competitor_starts_from_the_bucket_picks_up_the_nearest_potato_runs_back_with_it_drops_it_in_the_bucket_runs_back_to_pick_up_the_next_potato_runs_to_the_bucket_to_drop_it_in_and_she_continues_in_the_same_way_until_all_the_potatoes_are_in_the_bucket_What_is_the_total_distance_the_competitor_has_to_run\"><\/span><strong>20. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 3\" style=\"height: 82px; width: 565px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image162.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The distance of first potato from the starting point = 5 meters<\/p>\n<p style=\"text-align: justify;\">Therefore, the distance covered by competitor to pick up first potato and put it in bucket = <img decoding=\"async\" style=\"height: 19px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image163.png\" \/>= 10 meters<\/p>\n<p style=\"text-align: justify;\">The distance of Second potato from the starting point = 5 + 3 = 8 meters {All the potatoes are 3 meters apart from each other}<\/p>\n<p style=\"text-align: justify;\">Therefore, the distance covered by competitor to pick up 2nd potato and put it in bucket = <img decoding=\"async\" style=\"height: 19px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image164.png\" \/>= 16 meters<\/p>\n<p style=\"text-align: justify;\">The distance of third potato from the starting point = 8 + 3 = 11 meters<\/p>\n<p style=\"text-align: justify;\">Therefore, the distance covered by competitor to pick up 3rd potato and put it in bucket = <img decoding=\"async\" style=\"height: 17px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image165.png\" \/>= 22 meters<\/p>\n<p style=\"text-align: justify;\">Therefore, we have a sequence of the form 10, 16, 22 &#8230; 10 terms<\/p>\n<p style=\"text-align: justify;\">(There are ten terms because there are ten potatoes)<\/p>\n<p style=\"text-align: justify;\">To calculate the total distance covered by the competitor, we need to find:<\/p>\n<p style=\"text-align: justify;\">10 + 16 + 22 + &#8230; 10 terms<\/p>\n<p style=\"text-align: justify;\">First term = a = 10, Common difference = d = 16 \u2013 10 = 6<\/p>\n<p style=\"text-align: justify;\">n = 10{There are total of 10 terms in the sequence}<\/p>\n<p style=\"text-align: justify;\">Applying formula, <img decoding=\"async\" style=\"height: 42px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image002.png\" \/>to find sum of n terms of AP, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 327px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.3\/image166.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, total distance covered by competitor is equal to 370 meters.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_53\"><\/span>NCERT Solutions for Class 10 Maths Exercise 5.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 5.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class &#8230; <a title=\"NCERT Solutions for Class 10 Maths Exercise 5.3\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\" aria-label=\"More on NCERT Solutions for Class 10 Maths Exercise 5.3\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":30015,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1347,281],"tags":[1042,283,438,321,1485,216],"class_list":["post-4956","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-ncert-solutions","tag-cbse-class-10-mathematics","tag-cbse-study-material","tag-class-10","tag-mathematics","tag-maths","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2016-05-19T04:19:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-03-22T09:49:38+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1\" \/>\n\t<meta property=\"og:image:height\" content=\"1\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"17 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"NCERT Solutions for Class 10 Maths Exercise 5.3\",\"datePublished\":\"2016-05-19T04:19:00+00:00\",\"dateModified\":\"2023-03-22T09:49:38+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\"},\"wordCount\":3423,\"commentCount\":15,\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg\",\"keywords\":[\"CBSE Class 10 Mathematics\",\"CBSE Study Material\",\"Class 10\",\"Mathematics\",\"maths\",\"NCERT Solutions\"],\"articleSection\":[\"Mathematics\",\"NCERT Solutions\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\",\"name\":\"NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg\",\"datePublished\":\"2016-05-19T04:19:00+00:00\",\"dateModified\":\"2023-03-22T09:49:38+00:00\",\"description\":\"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics\",\"breadcrumb\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg\",\"caption\":\"NCERT Solutions for Class 10 Maths Exercise 5.3\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mycbseguide.com\/blog\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"NCERT Solutions for Class 10 Maths Exercise 5.3\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"name\":\"myCBSEguide\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\",\"name\":\"myCBSEguide\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"width\":180,\"height\":180,\"caption\":\"myCBSEguide\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/mycbseguide\/\",\"https:\/\/x.com\/mycbseguide\",\"https:\/\/www.linkedin.com\/company\/mycbseguide\/\",\"http:\/\/in.pinterest.com\/mycbseguide\/\",\"https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\",\"name\":\"myCBSEguide\",\"sameAs\":[\"http:\/\/mycbseguide.com\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide","description":"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide","og_description":"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics","og_url":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/","og_site_name":"myCBSEguide","article_publisher":"https:\/\/www.facebook.com\/mycbseguide\/","article_published_time":"2016-05-19T04:19:00+00:00","article_modified_time":"2023-03-22T09:49:38+00:00","og_image":[{"url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg","width":1,"height":1,"type":"image\/jpeg"}],"author":"myCBSEguide","twitter_card":"summary_large_image","twitter_creator":"@mycbseguide","twitter_site":"@mycbseguide","twitter_misc":{"Written by":"myCBSEguide","Est. reading time":"17 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#article","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/"},"author":{"name":"myCBSEguide","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65"},"headline":"NCERT Solutions for Class 10 Maths Exercise 5.3","datePublished":"2016-05-19T04:19:00+00:00","dateModified":"2023-03-22T09:49:38+00:00","mainEntityOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/"},"wordCount":3423,"commentCount":15,"publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage"},"thumbnailUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg","keywords":["CBSE Class 10 Mathematics","CBSE Study Material","Class 10","Mathematics","maths","NCERT Solutions"],"articleSection":["Mathematics","NCERT Solutions"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/","url":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/","name":"NCERT Solutions for Class 10 Maths Exercise 5.3 | myCBSEguide","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage"},"thumbnailUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg","datePublished":"2016-05-19T04:19:00+00:00","dateModified":"2023-03-22T09:49:38+00:00","description":"NCERT Solutions for Class 10 Maths Exercise 5.3 in PDF format for free download. NCERT Solutions Class 10 Mathematics","breadcrumb":{"@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#primaryimage","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/;https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10_english_ncert_solutions.jpg","caption":"NCERT Solutions for Class 10 Maths Exercise 5.3"},{"@type":"BreadcrumbList","@id":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-3\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mycbseguide.com\/blog\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/"},{"@type":"ListItem","position":3,"name":"NCERT Solutions for Class 10 Maths Exercise 5.3"}]},{"@type":"WebSite","@id":"https:\/\/mycbseguide.com\/blog\/#website","url":"https:\/\/mycbseguide.com\/blog\/","name":"myCBSEguide","description":"","publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/mycbseguide.com\/blog\/#organization","name":"myCBSEguide","url":"https:\/\/mycbseguide.com\/blog\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","width":180,"height":180,"caption":"myCBSEguide"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/mycbseguide\/","https:\/\/x.com\/mycbseguide","https:\/\/www.linkedin.com\/company\/mycbseguide\/","http:\/\/in.pinterest.com\/mycbseguide\/","https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ"]},{"@type":"Person","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65","name":"myCBSEguide","sameAs":["http:\/\/mycbseguide.com"]}]}},"_links":{"self":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/4956","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/comments?post=4956"}],"version-history":[{"count":3,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/4956\/revisions"}],"predecessor-version":[{"id":16221,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/4956\/revisions\/16221"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media\/30015"}],"wp:attachment":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media?parent=4956"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/categories?post=4956"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/tags?post=4956"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}