{"id":4952,"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-5-2\/"},"modified":"2023-03-22T15:18:38","modified_gmt":"2023-03-22T09:48:38","slug":"ncert-solutions-class-10-maths-exercise-5-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 5.2"},"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-5-2\/#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-5-2\/#1_Find_the_missing_variable_from_a_d_n_and_an_where_a_is_the_first_term_d_is_the_common_difference_and_an_is_the_nth_term_of_AP\" >1. Find the missing variable from a, d, n and an, where a is the first term, d is the common difference and an is the nth term of AP.<\/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-5-2\/#2Choose_the_correct_choice_in_the_following_and_justify\" >2.Choose the correct choice in the following and justify:<\/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-5-2\/#3_In_the_following_APs_find_the_missing_terms\" >3. In the following AP&#8217;s find the missing terms:<\/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-5-2\/#4_Which_term_of_the_AP_3_8_13_18_%E2%80%A6_is_78\" >4. Which term of the AP: 3, 8, 13, 18 &#8230; is 78?<\/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-5-2\/#6_Check_whether_%E2%80%93150_is_a_term_of_the_AP_11_8_5_2%E2%80%A6\" >6. Check whether \u2013150 is a term of the AP: 11, 8, 5, 2&#8230;<\/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-5-2\/#7_Find_the_31st_term_of_an_AP_whose_11th_term_is_38_and_16th_term_is_73\" >7. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.<\/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-5-2\/#8_An_AP_consists_of_50_terms_of_which_3rd_term_is_12_and_the_last_term_is_106_Find_the_29th_term\" >8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.<\/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-5-2\/#9_If_the_third_and_the_ninth_terms_of_an_AP_are_4_and_%E2%80%938_respectively_which_term_of_this_AP_is_zero\" >9. If the third and the ninth terms of an AP are 4 and \u20138 respectively, which term of this AP is zero?<\/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-5-2\/#10_The_17th_term_of_an_AP_exceeds_its_10th_term_by_7_Find_the_common_difference\" >10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.<\/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-5-2\/#11_Which_term_of_the_AP_3_15_27_39%E2%80%A6_will_be_132_more_than_its_54th_term\" >11. Which term of the AP: 3, 15, 27, 39&#8230; will be 132 more than its 54th term?<\/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-5-2\/#12_Two_APs_have_the_same_common_difference_The_difference_between_their_100th_terms_is_100_what_is_the_difference_between_their_1000th_terms\" >12. Two AP&#8217;s have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms.<\/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-5-2\/#13_How_many_three_digit_numbers_are_divisible_by_7\" >13. How many three digit numbers are divisible by 7?<\/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-5-2\/#14_How_many_multiples_of_4_lie_between_10_and_250\" >14. How many multiples of 4 lie between 10 and 250?<\/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-5-2\/#15_For_what_value_of_n_are_the_nth_terms_of_two_APs_63_65_67%E2%80%A6_and_3_10_17%E2%80%A6_equal\" >15. For what value of n, are the nth terms of two AP&#8217;s: 63, 65, 67&#8230; and 3, 10, 17&#8230; equal?<\/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-5-2\/#16_Determine_the_AP_whose_third_term_is_16_and_the_7th_term_exceeds_the_5th_term_by_12\" >16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.<\/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-5-2\/#17_Find_the_20th_term_from_the_last_term_of_the_AP_3_8_13%E2%80%A6_253\" >17. Find the 20th term from the last term of the AP: 3, 8, 13&#8230; , 253.<\/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-5-2\/#18_The_sum_of_the_4th_and_8th_terms_of_an_AP_is_24_and_the_sum_of_6th_and_10th_terms_is_44_Find_the_three_terms_of_the_AP\" >18. The sum of the 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the three terms of the AP.<\/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-5-2\/#19_Subba_Rao_started_work_in_1995_at_an_annual_salary_of_Rs_5000_and_received_an_increment_of_Rs_200_each_year_In_which_year_did_his_income_reach_Rs_7000\" >19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-2\/#20_Ramkali_saved_Rs_5_in_the_first_week_of_a_year_and_then_increased_her_weekly_savings_by_Rs_175_If_in_the_nth_week_her_weekly_savings_become_Rs_2075_find_n\" >20. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.<\/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-21\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-2\/#NCERT_Solutions_for_Class_10_Maths_Exercise_52\" >NCERT Solutions for Class 10 Maths Exercise 5.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-22\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-2\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 5.2 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 5.2\" width=\"181\" height=\"224\" \/><\/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_missing_variable_from_a_d_n_and_an_where_a_is_the_first_term_d_is_the_common_difference_and_an_is_the_nth_term_of_AP\"><\/span><strong>1. Find the missing variable from a, d, n and <em>a<sub>n<\/sub><\/em>, where a is the first term, d is the common difference and <em>a<sub>n<\/sub> <\/em>is the nth term of AP.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) a = 7, d = 3, n = 8<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) a = \u201318, n = 10, <img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image001.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) d = \u20133, n = 18, <img decoding=\"async\" style=\"height: 24px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image002.png\" \/> <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) a = \u201318.9, d = 2.5, <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image003.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(v) a = 3.5, d = 0, n = 105<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> a = 7, d = 3, n = 8<\/p>\n<p style=\"text-align: justify;\">We need to find <img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image004.png\" \/> here.<\/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\/Ex5.2\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting values of a, d and n,<\/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\/Ex5.2\/image006.png\" \/>= 7 + (8 \u2212 1) 3<\/p>\n<p style=\"text-align: justify;\">= 7 + (7) 3 = 7 + 21 = 28<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> a = \u201318, n= 10, <img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">We need to find d here.<\/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\/Ex5.2\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting values of a, <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image008.png\" \/>,<\/p>\n<p style=\"text-align: justify;\">0 = \u201318 + (10 \u2013 1) d<\/p>\n<p style=\"text-align: justify;\">\u21d2 0 = \u221218 + 9<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 18 = 9<em>d <\/em>\u21d2 <em>d <\/em>= 2<\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> d = \u20133, n = 18, <img decoding=\"async\" style=\"height: 24px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image009.png\" \/><\/p>\n<p style=\"text-align: justify;\">We need to find a here.<\/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\/Ex5.2\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting values of d, <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image008.png\" \/>,<\/p>\n<p style=\"text-align: justify;\">\u20135 = a + (18 \u2013 1) (\u20133)<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22125 = <em>a <\/em>+ (17) (\u22123)<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22125 = <em>a <\/em>\u2013 51 \u21d2 <em>a <\/em>= 46<\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<\/strong> a = \u201318.9, d = 2.5, <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\">We need to find n here.<\/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\/Ex5.2\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting values of d, <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image008.png\" \/>,<\/p>\n<p style=\"text-align: justify;\">3.6 = \u201318.9 + (n \u2013 1) (2.5)<\/p>\n<p style=\"text-align: justify;\">\u21d2 3.6 = \u221218.9 + 2.5<em>n <\/em>\u2212 2.5<\/p>\n<p style=\"text-align: justify;\">\u21d2 2.5<em>n <\/em>= 25 \u21d2 <em>n <\/em>= 10<\/p>\n<p style=\"text-align: justify;\"><strong>(v)<\/strong> a = 3.5, d = 0, n = 105<\/p>\n<p style=\"text-align: justify;\">We need to find <img decoding=\"async\" style=\"height: 24px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image004.png\" \/> here.<\/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\/Ex5.2\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting values of d, n and a,<\/p>\n<p style=\"text-align: justify;\"><em>a<sub>n<\/sub><\/em> = 3.5 + (105 \u2212 1) (0)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image011.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image012.png\" \/>\u21d2 <img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image013.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2Choose_the_correct_choice_in_the_following_and_justify\"><\/span><strong>2.Choose the correct choice in the following and justify:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) 30<sup>th<\/sup> term of the AP: 10, 7, 4&#8230; is<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(A) 97 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(B) 77 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(C) \u201377 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(D) \u201387<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) 11<sup>th<\/sup> term of the AP: \u22123, \u221212, 2&#8230; is<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(A) 28 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(B) 22 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(C) \u201338 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(D) <img decoding=\"async\" style=\"height: 42px; width: 44px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image014.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans.(i)<\/strong> 10, 7, 4&#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 10, Common difference = d = 7 \u2013 10 = 4 \u2013 7 = \u20133<\/p>\n<p style=\"text-align: justify;\">And n = 30{Because, we need to find 30<sup>th<\/sup> term}<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image005.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\/Ex5.2\/image015.png\" \/>= 10 + (30 \u2212 1) (\u22123) = 10 \u2013 87 = \u221277<\/p>\n<p style=\"text-align: justify;\">Therefore, the answer is (C).<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> \u22123, \u2212\u00bd, 2&#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = \u20133, Common difference = d = \u2212<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image016.png\" \/> \u2212 (\u22123) = 2 \u2212 (\u2212<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image016.png\" \/>) = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image017.png\" \/><\/p>\n<p style=\"text-align: justify;\">And n = 11 (Because, we need to find 11<sup>th<\/sup> term)<\/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\/Ex5.2\/image006.png\" \/>= \u22123 + (11 \u2013 1) <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image017.png\" \/>= \u22123 + 25 = 22<\/p>\n<p style=\"text-align: justify;\">Therefore 11<sup>th<\/sup> term is 22 which means answer is (B).<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_In_the_following_APs_find_the_missing_terms\"><\/span><strong>3. In the following AP&#8217;s find the missing terms:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) 2, __ , 26 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) __, 13, __, 3<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) 5, __, __, <img decoding=\"async\" style=\"height: 42px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image018.png\" \/> <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) \u20134. __, __, __, __, 6<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(v) __, 38, __, __, __, \u201322<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> 2, __ , 26<\/p>\n<p style=\"text-align: justify;\">We know that difference between consecutive terms is equal in any A.P.<\/p>\n<p style=\"text-align: justify;\">Let the missing term be <em>x<\/em>.<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>\u2013 2 = 26 \u2013 <em>x<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>x <\/em>= 28 \u21d2 <em>x <\/em>= 14<\/p>\n<p style=\"text-align: justify;\">Therefore, missing term is 14.<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong>__, 13, __, 3<\/p>\n<p style=\"text-align: justify;\">Let missing terms be <em>x<\/em> and <em>y<\/em>.<\/p>\n<p style=\"text-align: justify;\">The sequence becomes <em>x<\/em>, 13, <em>y<\/em>, 3<\/p>\n<p style=\"text-align: justify;\">We know that difference between consecutive terms is constant in any A.P.<\/p>\n<p style=\"text-align: justify;\"><em>y <\/em>\u2013 13 = 3 \u2013 <em>y<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>y <\/em>= 16 \u21d2 <em>y <\/em>= 8<\/p>\n<p style=\"text-align: justify;\">And 13 \u2013 <em>x <\/em>= <em>y <\/em>\u2013 13<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>+ <em>y <\/em>= 26<\/p>\n<p style=\"text-align: justify;\">But, we have <em>y <\/em>= 8,<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>+ 8 = 26\u21d2 <em>x <\/em>= 18<\/p>\n<p style=\"text-align: justify;\">Therefore, missing terms are 18 and 8.<\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> 5, __, __, <img decoding=\"async\" style=\"height: 42px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here, first term = <em>a <\/em>= 5 And, 4<sup>th<\/sup> term = <img decoding=\"async\" style=\"height: 42px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image019.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\/Ex5.2\/image005.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\/Ex5.2\/image020.png\" \/>= 5 + (4 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image021.png\" \/>= 5 + 3<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 19 = 2 (5 + 3d)<\/p>\n<p style=\"text-align: justify;\">\u21d2 19 = 10 + 6d<\/p>\n<p style=\"text-align: justify;\">\u21d2 6d = 19 \u2013 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 6d = 9\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\/Ex5.2\/image022.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, we get common difference = d = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image022.png\" \/><\/p>\n<p style=\"text-align: justify;\">Second term = a + d = <img decoding=\"async\" style=\"height: 42px; width: 69px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image023.png\" \/><\/p>\n<p style=\"text-align: justify;\">Third term = second term + d = <img decoding=\"async\" style=\"height: 42px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image024.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, missing terms are <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image025.png\" \/>and 8<\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<\/strong>\u20134. __, __, __, __, 6<\/p>\n<p style=\"text-align: justify;\">Here, First term = a = \u20134 and 6<sup>th<\/sup> term = <img decoding=\"async\" style=\"height: 24px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image026.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\/Ex5.2\/image005.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\/Ex5.2\/image027.png\" \/>= \u22124 + (6 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 6 = \u22124 + 5<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 5<em>d <\/em>= 10 \u21d2 <em>d <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Therefore, common difference = d = 2<\/p>\n<p style=\"text-align: justify;\">Second term = first term + d = a + d = \u20134 + 2 = \u20132<\/p>\n<p style=\"text-align: justify;\">Third term = second term + d = \u20132 + 2 = 0<\/p>\n<p style=\"text-align: justify;\">Fourth term = third term + d = 0 + 2 = 2<\/p>\n<p style=\"text-align: justify;\">Fifth term = fourth term + d = 2 + 2 = 4<\/p>\n<p style=\"text-align: justify;\">Therefore, missing terms are \u20132, 0, 2 and 4.<\/p>\n<p style=\"text-align: justify;\"><strong>(v)<\/strong> __, 38, __, __, __, \u201322<\/p>\n<p style=\"text-align: justify;\">We are given 2<sup>nd<\/sup> and 6<sup>th<\/sup> term.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a<\/em><sub>2<\/sub> = <em>a <\/em>+ (2 \u2212 1) <em>d <\/em>And <em>a<\/em><sub>6<\/sub> = <em>a <\/em>+ (6 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 38 = <em>a <\/em>+ <em>d <\/em>And \u221222 = <em>a <\/em>+ 5<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">These are equations in two variables, we can solve them using any method.<\/p>\n<p style=\"text-align: justify;\">Using equation (38 = <em>a <\/em>+ <em>d<\/em>), we can say that <em>a <\/em>= 38 \u2212 <em>d<\/em>.<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>a<\/em> in equation (\u221222 = <em>a <\/em>+ 5<em>d<\/em>),<\/p>\n<p style=\"text-align: justify;\">\u221222 = 38 \u2013 <em>d <\/em>+ 5<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 4<em>d <\/em>= \u221260<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>d <\/em>= \u221215<\/p>\n<p style=\"text-align: justify;\">Using this value of <em>d<\/em> and putting this in equation 38 = <em>a <\/em>+ <em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">38 = <em>a <\/em>\u2013 15\u21d2 <em>a <\/em>= 53<\/p>\n<p style=\"text-align: justify;\">Therefore, we get <em>a <\/em>= 53 and <em>d <\/em>= \u221215<\/p>\n<p style=\"text-align: justify;\">First term = a = 53<\/p>\n<p style=\"text-align: justify;\">Third term = second term + d = 38 \u2013 15 = 23<\/p>\n<p style=\"text-align: justify;\">Fourth term = third term + d = 23 \u2013 15 = 8<\/p>\n<p style=\"text-align: justify;\">Fifth term = fourth term + d = 8 \u2013 15 = \u20137<\/p>\n<p style=\"text-align: justify;\">Therefore, missing terms are 53, 23, 8 and \u20137.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Which_term_of_the_AP_3_8_13_18_%E2%80%A6_is_78\"><\/span><strong>4. Which term of the AP: 3, 8, 13, 18 &#8230; is 78?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>First term = <em>a <\/em>= 3, Common difference = d = 8 \u2013 3 = 13 \u2013 8 = 5 and <em>a<sub>n<\/sub><\/em> = 78<\/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\/Ex5.2\/image005.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\/Ex5.2\/image006.png\" \/>= 3 + (<em>n <\/em>\u2212 1) 5,<\/p>\n<p style=\"text-align: justify;\">\u21d2 78 = 3 + (<em>n <\/em>\u2212 1) 5<\/p>\n<p style=\"text-align: justify;\">\u21d2 75 = 5<em>n <\/em>\u2212 5<\/p>\n<p style=\"text-align: justify;\">\u21d2 80 = 5<em>n<\/em>\u21d2 <em>n <\/em>= 16<\/p>\n<p style=\"text-align: justify;\">It means 16<sup>th<\/sup> term of the given AP is equal to 78.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<p style=\"text-align: justify;\"><strong>5. Find the number of terms in each of the following APs:<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(i) 7, 13, 19&#8230;., 205<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) 18, <img decoding=\"async\" style=\"height: 42px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image028.png\" \/>, 13&#8230;, \u221247<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> 7, 13, 19 &#8230;, 205<\/p>\n<p style=\"text-align: justify;\">First term = a = 7, Common difference = d = 13 \u2013 7 = 19 \u2013 13 = 6<\/p>\n<p style=\"text-align: justify;\">And <img decoding=\"async\" style=\"height: 24px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image029.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\/Ex5.2\/image005.png\" \/>, to find nth term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">205 = 7 + (<em>n <\/em>\u2212 1) 6 = 7 + 6<em>n <\/em>\u2013 6<\/p>\n<p style=\"text-align: justify;\">\u21d2 205 = 6<em>n <\/em>+ 1<\/p>\n<p style=\"text-align: justify;\">\u21d2 204 = 6<em>n<\/em>\u21d2 <em>n <\/em>= 34<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 34 terms in the given arithmetic progression.<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> 18, <img decoding=\"async\" style=\"height: 42px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image028.png\" \/>, 13 &#8230;, \u221247<\/p>\n<p style=\"text-align: justify;\">First term = a =18, Common difference = d = <img decoding=\"async\" style=\"height: 42px; width: 218px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image030.png\" \/><\/p>\n<p style=\"text-align: justify;\">And <img decoding=\"async\" style=\"height: 24px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image031.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\/Ex5.2\/image005.png\" \/>, to find nth term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">\u221247 = 18 + (<em>n <\/em>\u2212 1) <img decoding=\"async\" style=\"height: 45px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image032.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 36 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image017.png\" \/><em>n <\/em>+ <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image017.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 \u221294 = 36 \u2212 5<em>n <\/em>+ 5<\/p>\n<p style=\"text-align: justify;\">\u21d2 5<em>n <\/em>= 135 \u21d2 <em>n <\/em>= 27<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 27 terms in the given arithmetic progression.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_Check_whether_%E2%80%93150_is_a_term_of_the_AP_11_8_5_2%E2%80%A6\"><\/span><strong>6. Check whether \u2013150 is a term of the AP: 11, 8, 5, 2&#8230;<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let \u2212150 is the n<sup>th<\/sup> of AP 11, 8, 5, 2&#8230; which means that <img decoding=\"async\" style=\"height: 24px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image033.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here, First term = a = 11, Common difference = d = 8 \u2013 11 = \u20133<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">\u2212150 = 11 + (<em>n <\/em>\u2212 1) (\u22123)<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u2212150 = 11 \u2212 3<em>n <\/em>+ 3<\/p>\n<p style=\"text-align: justify;\">\u21d2 3<em>n <\/em>= 164 \u21d2 <em>n <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 29px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image034.png\" \/><\/p>\n<p style=\"text-align: justify;\">But, n cannot be in fraction.<\/p>\n<p style=\"text-align: justify;\">Therefore, our supposition is wrong. \u2212150 cannot be term in AP.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_Find_the_31st_term_of_an_AP_whose_11th_term_is_38_and_16th_term_is_73\"><\/span><strong>7. Find the 31<sup>st<\/sup> term of an AP whose 11<sup>th<\/sup> term is 38 and 16<sup>th<\/sup> term is 73.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Here <img decoding=\"async\" style=\"height: 24px; width: 143px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image035.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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">38 = <em>a <\/em>+ (11 \u2212 1) (<em>d<\/em>)And73 = <em>a <\/em>+ (16 \u2212 1) (<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 38 = <em>a <\/em>+ 10<em>d <\/em>And 73 = <em>a <\/em>+ 15<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">These are equations consisting of two variables.<\/p>\n<p style=\"text-align: justify;\">We have, 38 = <em>a <\/em>+ 10<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= 38 \u2212 10<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Let us put value of a in equation (73 = <em>a <\/em>+ 15<em>d<\/em>),<\/p>\n<p style=\"text-align: justify;\">73 = 38 \u2212 10<em>d <\/em>+ 15<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 35 = 5<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Therefore, Common difference = <em>d <\/em>= 7<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in equation 38 = <em>a <\/em>+ 10<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">38 = <em>a <\/em>+ 70<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>= \u221232<\/p>\n<p style=\"text-align: justify;\">Therefore, common difference = d = 7 and First term = a = \u201332<\/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\/Ex5.2\/image005.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\/Ex5.2\/image036.png\" \/>= \u221232 + (31 \u2212 1) (7)<\/p>\n<p style=\"text-align: justify;\">= \u221232 + 210 = 178<\/p>\n<p style=\"text-align: justify;\">Therefore, 31<sup>st<\/sup> term of AP is 178.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_An_AP_consists_of_50_terms_of_which_3rd_term_is_12_and_the_last_term_is_106_Find_the_29th_term\"><\/span><strong>8. An AP consists of 50 terms of which 3<sup>rd<\/sup> term is 12 and the last term is 106. Find the 29<sup>th<\/sup> term.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>An AP consists of 50 terms and the 50<sup>th<\/sup> term is equal to 106 and <img decoding=\"async\" style=\"height: 24px; width: 54px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image037.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\/Ex5.2\/image005.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: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image038.png\" \/>= <em>a <\/em>+ (50 \u2212 1) <em>d <\/em>And <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image039.png\" \/>= <em>a <\/em>+ (3 \u2212 1) <em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 106 = <em>a <\/em>+ 49<em>d <\/em>And12 = <em>a <\/em>+ 2<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">These are equations consisting of two variables.<\/p>\n<p style=\"text-align: justify;\">Using equation 106 = <em>a <\/em>+ 49<em>d<\/em>, we get <em>a <\/em>= 106 \u2212 49<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting value of <em>a<\/em> in the equation 12 = <em>a <\/em>+ 2<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">12 = 106 \u2212 49<em>d <\/em>+ 2<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 47<em>d <\/em>= 94 \u21d2 <em>d <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in the equation, <em>a <\/em>= 106 \u2212 49<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>= 106 \u2013 49 (2) = 106 \u2013 98 = 8<\/p>\n<p style=\"text-align: justify;\">Therefore, First term = <em>a <\/em>= 8 and Common difference = <em>d <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">To find 29<sup>th<\/sup> term, we use formula <img decoding=\"async\" style=\"height: 26px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image005.png\" \/> which is used to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image040.png\" \/>= 8 + (29 \u2212 1) 2 = 8 + 56 = 64<\/p>\n<p style=\"text-align: justify;\">Therefore, 29th term of AP is equal to 64.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_If_the_third_and_the_ninth_terms_of_an_AP_are_4_and_%E2%80%938_respectively_which_term_of_this_AP_is_zero\"><\/span><strong>9. If the third and the ninth terms of an AP are 4 and \u20138 respectively, which term of this AP is zero?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is given that 3<sup>rd<\/sup> and 9<sup>th<\/sup> term of AP are 4 and \u20138 respectively.<\/p>\n<p style=\"text-align: justify;\">It means <img decoding=\"async\" style=\"height: 24px; width: 135px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image041.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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">4 = a + (3 \u2013 1) d And, \u20138 = a + (9 \u2013 1) d<\/p>\n<p style=\"text-align: justify;\">\u21d2 4 = <em>a <\/em>+ 2<em>d <\/em>And, \u22128 = <em>a <\/em>+ 8<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">These are equations in two variables.<\/p>\n<p style=\"text-align: justify;\">Using equation 4 = <em>a <\/em>+ 2<em>d<\/em>, we can say that <em>a <\/em>= 4 \u2212 2<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">Putting value of <em>a<\/em> in other equation \u22128 = <em>a <\/em>+ 8<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">\u22128 = 4 \u2212 2<em>d <\/em>+ 8<em>d<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 \u221212 = 6<em>d <\/em>\u21d2 <em>d <\/em>= \u22122<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in equation \u22128 = <em>a <\/em>+ 8<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">\u22128 = <em>a <\/em>+ 8 (\u22122)<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u22128 = <em>a <\/em>\u2013 16 \u21d2 <em>a <\/em>= 8<\/p>\n<p style=\"text-align: justify;\">Therefore, first term = <em>a <\/em>= 8 and Common Difference = <em>d <\/em>= \u22122<\/p>\n<p style=\"text-align: justify;\">We want to know which term is equal to zero.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">0 = 8 + (<em>n <\/em>\u2212 1) (\u22122)<\/p>\n<p style=\"text-align: justify;\">\u21d2 0 = 8 \u2212 2<em>n <\/em>+ 2<\/p>\n<p style=\"text-align: justify;\">\u21d2 0 = 10 \u2212 2<em>n<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>n <\/em>= 10\u21d2 <em>n <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Therefore, 5<sup>th<\/sup> term is equal to 0.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"10_The_17th_term_of_an_AP_exceeds_its_10th_term_by_7_Find_the_common_difference\"><\/span><strong>10. The 17<sup>th<\/sup> term of an AP exceeds its 10<sup>th<\/sup> term by 7. Find the common difference.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <img decoding=\"async\" style=\"height: 24px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image042.png\" \/><\/strong>(1)<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a<\/em><sub>17<\/sub> = <em>a <\/em>+ 16<em>d\u2026<\/em> (2)<\/p>\n<p style=\"text-align: justify;\"><em>a<\/em><sub>10<\/sub> = <em>a <\/em>+ 9<em>d\u2026 <\/em>(3)<\/p>\n<p style=\"text-align: justify;\">Putting (2) and (3) in equation (1),<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>+ 16<em>d <\/em>= <em>a <\/em>+ 9<em>d <\/em>+ 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 7<em>d <\/em>= 7\u21d2 <em>d <\/em>= 1<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"11_Which_term_of_the_AP_3_15_27_39%E2%80%A6_will_be_132_more_than_its_54th_term\"><\/span><strong>11. Which term of the AP: 3, 15, 27, 39&#8230; will be 132 more than its 54<sup>th<\/sup> term?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Lets first calculate 54<sup>th<\/sup> of the given AP.<\/p>\n<p style=\"text-align: justify;\">First term = a = 3, Common difference = d = 15 \u2013 3 = 12<\/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\/Ex5.2\/image005.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: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image043.png\" \/>= <em>a <\/em>+ (54 \u2212 1) <em>d <\/em>= 3 + 53 (12) = 3 + 636 = 639<\/p>\n<p style=\"text-align: justify;\">We want to find which term is 132 more than its 54<sup>th<\/sup> term.<\/p>\n<p style=\"text-align: justify;\">Let us suppose it is n<sup>th<\/sup> term which is 132 more than 54<sup>th<\/sup> term.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image044.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 3 + (<em>n <\/em>\u2212 1) 12 = 639 + 132<\/p>\n<p style=\"text-align: justify;\">\u21d2 3 + 12<em>n <\/em>\u2013 12 = 771<\/p>\n<p style=\"text-align: justify;\">\u21d2 12<em>n <\/em>\u2013 9 = 771<\/p>\n<p style=\"text-align: justify;\">\u21d2 12<em>n <\/em>= 780\u21d2 <em>n <\/em>= 65<\/p>\n<p style=\"text-align: justify;\">Therefore, 65<sup>th<\/sup> term is 132 more than its 54<sup>th<\/sup> term.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"12_Two_APs_have_the_same_common_difference_The_difference_between_their_100th_terms_is_100_what_is_the_difference_between_their_1000th_terms\"><\/span><strong>12. Two AP&#8217;s have the same common difference. The difference between their 100<sup>th<\/sup> terms is 100, what is the difference between their 1000<sup>th<\/sup> terms.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let first term of 1<sup>st<\/sup> AP = <em>a<\/em><\/p>\n<p style=\"text-align: justify;\">Let first term of 2<sup>nd<\/sup> AP = <em>a<\/em>\u2032<\/p>\n<p style=\"text-align: justify;\">It is given that their common difference is same.<\/p>\n<p style=\"text-align: justify;\">Let their common difference be <em>d<\/em>.<\/p>\n<p style=\"text-align: justify;\">It is given that difference between their 100<sup>th<\/sup> terms is 100.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>+ (100 \u2212 1) <em>d <\/em>\u2013 [<em>a<\/em>\u2032 + (100 \u2212 1) <em>d<\/em>]\n<p style=\"text-align: justify;\">= <em>a <\/em>+ 99<em>d <\/em>\u2212 <em>a<\/em>\u2032 \u2212 99<em>d <\/em>= 100<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>\u2212 <em>a<\/em>\u2032 = 100\u2026 (1)<\/p>\n<p style=\"text-align: justify;\">We want to find difference between their 1000<sup>th<\/sup> terms which means we want to calculate:<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>+ (1000 \u2212 1) <em>d <\/em>\u2013 [<em>a<\/em>\u2032 + (1000 \u2212 1) <em>d<\/em>]\n<p style=\"text-align: justify;\">= <em>a <\/em>+ 999<em>d <\/em>\u2212 <em>a<\/em>\u2032 \u2212 999<em>d <\/em>= <em>a <\/em>\u2013 <em>a<\/em>\u2032<\/p>\n<p style=\"text-align: justify;\">Putting equation (1) in the above equation,<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>+ (1000 \u2212 1) <em>d <\/em>\u2013 [<em>a<\/em>\u2032 + (1000 \u2212 1) <em>d<\/em>]\n<p style=\"text-align: justify;\">= <em>a <\/em>+ 999<em>d <\/em>\u2212 <em>a<\/em>\u2032 + 999<em>d <\/em>= <em>a <\/em>\u2212 <em>a<\/em>\u2032 = 100<\/p>\n<p style=\"text-align: justify;\">Therefore, difference between their 1000<sup>th<\/sup> terms would be equal to 100.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"13_How_many_three_digit_numbers_are_divisible_by_7\"><\/span><strong>13. How many three digit numbers are divisible by 7?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>We have AP starting from 105 because it is the first three digit number divisible by 7.<\/p>\n<p style=\"text-align: justify;\">AP will end at 994 because it is the last three digit number divisible by 7.<\/p>\n<p style=\"text-align: justify;\">Therefore, we have AP of the form 105, 112, 119&#8230;, 994<\/p>\n<p style=\"text-align: justify;\">Let 994 is the n<sup>th<\/sup> term of AP.<\/p>\n<p style=\"text-align: justify;\">We need to find n here.<\/p>\n<p style=\"text-align: justify;\">First term = a = 105, Common difference = d = 112 \u2013 105 = 7<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">994 = 105 + (<em>n <\/em>\u2212 1) (7)<\/p>\n<p style=\"text-align: justify;\">\u21d2 994 = 105 + 7<em>n <\/em>\u2212 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 896 = 7<em>n <\/em>\u21d2 <em>n <\/em>= 128<\/p>\n<p style=\"text-align: justify;\">It means 994 is the 128<sup>th<\/sup> term of AP.<\/p>\n<p style=\"text-align: justify;\">Therefore, there are 128 terms in AP.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"14_How_many_multiples_of_4_lie_between_10_and_250\"><\/span><strong>14. How many multiples of 4 lie between 10 and 250?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>First multiple of 4 which lie between 10 and 250 is 12.<\/p>\n<p style=\"text-align: justify;\">The last multiple of 4 which lie between 10 and 250 is 248.<\/p>\n<p style=\"text-align: justify;\">Therefore, AP is of the form 12, 16, 20&#8230; ,248<\/p>\n<p style=\"text-align: justify;\">First term = a = 12, Common difference = d = 4<\/p>\n<p style=\"text-align: justify;\">Using formula <em>a<sub>n<\/sub><\/em> = <em>a <\/em>+ (<em>n <\/em>\u2013 1) <em>d<\/em>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">248 = 12 + (<em>n <\/em>\u2212 1) (4)<\/p>\n<p style=\"text-align: justify;\">\u21d2 248 = 12 + 4<em>n <\/em>\u2212 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 240 = 4<em>n <\/em>\u21d2 <em>n <\/em>= 60<\/p>\n<p style=\"text-align: justify;\">It means that 248 is the 60<sup>th<\/sup> term of AP.<\/p>\n<p style=\"text-align: justify;\">So, we can say that there are 60 multiples of 4 which lie between 10 and 250.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"15_For_what_value_of_n_are_the_nth_terms_of_two_APs_63_65_67%E2%80%A6_and_3_10_17%E2%80%A6_equal\"><\/span><strong>15. For what value of n, are the nth terms of two AP&#8217;s: 63, 65, 67&#8230; and 3, 10, 17&#8230; equal?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Lets first consider AP 63, 65, 67&#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 63, Common difference = d = 65 \u2013 63 = 2<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a<sub>n<\/sub><\/em> = 63 + (<em>n <\/em>\u2212 1) (2)\u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Now, consider second AP 3, 10, 17&#8230;<\/p>\n<p style=\"text-align: justify;\">First term = a = 3, Common difference = d = 10 \u2013 3 = 7<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a<sub>n<\/sub><\/em> = 3 + (<em>n <\/em>\u2212 1) (7)\u2026 (2)<\/p>\n<p style=\"text-align: justify;\">According to the given condition:<\/p>\n<p style=\"text-align: justify;\">(1) = (2)<\/p>\n<p style=\"text-align: justify;\">\u21d2 63 + (<em>n <\/em>\u2212 1) (2) = 3 + (<em>n <\/em>\u2212 1) (7)<\/p>\n<p style=\"text-align: justify;\">\u21d2 63 + 2<em>n <\/em>\u2013 2 = 3 + 7<em>n <\/em>\u2212 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 65 = 5<em>n<\/em>\u21d2 <em>n <\/em>= 13<\/p>\n<p style=\"text-align: justify;\">Therefore, 13<sup>th<\/sup> terms of both the AP&#8217;s are equal.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"16_Determine_the_AP_whose_third_term_is_16_and_the_7th_term_exceeds_the_5th_term_by_12\"><\/span><strong>16. Determine the AP whose third term is 16 and the 7<sup>th<\/sup> term exceeds the 5<sup>th<\/sup> term by 12.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let first term of AP = a<\/p>\n<p style=\"text-align: justify;\">Let common difference of AP = d<\/p>\n<p style=\"text-align: justify;\">It is given that its 3<sup>rd<\/sup> term is equal to 16.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">16 = <em>a <\/em>+ (3 \u2212 1) (<em>d<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 16 = <em>a <\/em>+ 2<em>d\u2026<\/em> (1)<\/p>\n<p style=\"text-align: justify;\">It is also given that 7<sup>th<\/sup> term exceeds 5<sup>th<\/sup> term by12.<\/p>\n<p style=\"text-align: justify;\">According to the given condition:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image045.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>+ (7 \u2013 1) <em>d <\/em>= <em>a <\/em>+ (5 \u2013 1) <em>d <\/em>+ 12<\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>d <\/em>= 12 \u21d2 <em>d <\/em>= 6<\/p>\n<p style=\"text-align: justify;\">Putting value of d in equation 16 = <em>a <\/em>+ 2<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\">16 = <em>a <\/em>+ 2 (6) \u21d2 <em>a <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">Therefore, first term = a = 4<\/p>\n<p style=\"text-align: justify;\">And, common difference = d = 6<\/p>\n<p style=\"text-align: justify;\">Therefore, AP is 4, 10, 16, 22&#8230;<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"17_Find_the_20th_term_from_the_last_term_of_the_AP_3_8_13%E2%80%A6_253\"><\/span><strong>17. Find the 20<sup>th<\/sup> term from the last term of the AP: 3, 8, 13&#8230; , 253.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>We want to find 20<sup>th<\/sup> term from the last term of given AP.<\/p>\n<p style=\"text-align: justify;\">So, let us write given AP in this way:253 &#8230; 13, 8, 3<\/p>\n<p style=\"text-align: justify;\">Here First term = a = 253, Common Difference = d = 8 \u2013 13 = \u20135<\/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\/Ex5.2\/image005.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: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image046.png\" \/>= 253 + (20 \u2212 1) (\u22125)<\/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\/Ex5.2\/image046.png\" \/>= 253 + 19 (\u22125) = 253 \u2013 95 = 158<\/p>\n<p style=\"text-align: justify;\">Therefore, the 20<sup>th<\/sup> term from the last term of given AP is 158.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"18_The_sum_of_the_4th_and_8th_terms_of_an_AP_is_24_and_the_sum_of_6th_and_10th_terms_is_44_Find_the_three_terms_of_the_AP\"><\/span><strong>18. The sum of the 4<sup>th<\/sup> and 8<sup>th<\/sup> terms of an AP is 24 and the sum of 6<sup>th<\/sup> and 10<sup>th<\/sup> terms is 44. Find the three terms of the AP.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The sum of 4<sup>th<\/sup> and 8<sup>th<\/sup> terms of an AP is 24 and sum of 6<sup>th<\/sup> and 10<sup>th<\/sup> terms is 44.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/image047.png\" \/>and <img decoding=\"async\" style=\"height: 24px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch05\/Ex5.2\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>+ (4 \u2212 1) <em>d <\/em>+ [<em>a <\/em>+ (8 \u2212 1) <em>d<\/em>] = 24<\/p>\n<p style=\"text-align: justify;\">And, <em>a <\/em>+ (6 \u2212 1) <em>d <\/em>+ [<em>a <\/em>+ (10 \u2212 1) <em>d<\/em>] = 44<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>+ 3<em>d <\/em>+ <em>a <\/em>+ 7<em>d <\/em><\/p>\n<p style=\"text-align: justify;\">= 24 And <em>a <\/em>+ 5<em>d <\/em>+ <em>a <\/em>+ 9<em>d <\/em>= 44<\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>a <\/em>+ 10<em>d <\/em>= 24 And 2<em>a <\/em>+ 14<em>d <\/em>= 44<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>a <\/em>+ 5<em>d <\/em>= 12 And <em>a <\/em>+ 7<em>d <\/em>=22<\/p>\n<p style=\"text-align: justify;\">These are equations in two variables.<\/p>\n<p style=\"text-align: justify;\">Using equation, <em>a <\/em>+ 5<em>d <\/em>= 12, we can say that <em>a <\/em>= 12 \u2212 5<em>d\u2026 <\/em>(1)<\/p>\n<p style=\"text-align: justify;\">Putting (1) in equation <em>a <\/em>+ 7<em>d <\/em>= 22,<\/p>\n<p style=\"text-align: justify;\">12 \u2212 5<em>d <\/em>+ 7<em>d <\/em>= 22<\/p>\n<p style=\"text-align: justify;\">\u21d2 12 + 2<em>d <\/em>= 22<\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>d <\/em>= 10 \u21d2 <em>d <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Putting value of <em>d<\/em> in equation <em>a <\/em>= 12 \u2212 5<em>d<\/em>,<\/p>\n<p style=\"text-align: justify;\"><em>a <\/em>= 12 \u2013 5 (5) = 12 \u2013 25 = \u221213<\/p>\n<p style=\"text-align: justify;\">Therefore, first term = <em>a <\/em>= \u221213 and, Common difference = <em>d <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Therefore, AP is \u201313, \u20138, \u20133, 2&#8230;<\/p>\n<p style=\"text-align: justify;\">Its first three terms are \u201313, \u20138 and \u20133.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"19_Subba_Rao_started_work_in_1995_at_an_annual_salary_of_Rs_5000_and_received_an_increment_of_Rs_200_each_year_In_which_year_did_his_income_reach_Rs_7000\"><\/span><strong>19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Subba Rao&#8217;s starting salary = Rs 5000<\/p>\n<p style=\"text-align: justify;\">It means, first term = a = 5000<\/p>\n<p style=\"text-align: justify;\">He gets an increment of Rs 200 after every year.<\/p>\n<p style=\"text-align: justify;\">Therefore, common difference = d = 200<\/p>\n<p style=\"text-align: justify;\">His salary after 1 year = 5000 + 200 = Rs 5200<\/p>\n<p style=\"text-align: justify;\">His salary after two years = 5200 + 200 = Rs 5400<\/p>\n<p style=\"text-align: justify;\">Therefore, it is an AP of the form 5000, 5200, 5400, 5600&#8230; , 7000<\/p>\n<p style=\"text-align: justify;\">We want to know in which year his income reaches Rs 7000.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">7000 = 5000 + (<em>n <\/em>\u2212 1) (200)<\/p>\n<p style=\"text-align: justify;\">\u21d2 7000 = 5000 + 200<em>n <\/em>\u2212 200<\/p>\n<p style=\"text-align: justify;\">\u21d2 7000 \u2013 5000 + 200 = 200<em>n<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 2200 = 200<em>n<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>n <\/em>= 11<\/p>\n<p style=\"text-align: justify;\">It means after 11 years, Subba Rao&#8217;s income would be Rs 7000.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 5.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"20_Ramkali_saved_Rs_5_in_the_first_week_of_a_year_and_then_increased_her_weekly_savings_by_Rs_175_If_in_the_nth_week_her_weekly_savings_become_Rs_2075_find_n\"><\/span><strong>20. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Ramkali saved Rs. 5 in the first week of year. It means first term = a = 5<\/p>\n<p style=\"text-align: justify;\">Ramkali increased her weekly savings by Rs 1.75.<\/p>\n<p style=\"text-align: justify;\">Therefore, common difference = d = Rs 1.75<\/p>\n<p style=\"text-align: justify;\">Money saved by Ramkali in the second week = a + d = 5 + 1.75 = Rs 6.75<\/p>\n<p style=\"text-align: justify;\">Money saved by Ramkali in the third week = 6.75 + 1.75 = Rs 8.5<\/p>\n<p style=\"text-align: justify;\">Therefore, it is an AP of the form:5, 6.75, 8.5 &#8230; , 20.75<\/p>\n<p style=\"text-align: justify;\">We want to know in which year her weekly savings become 20.75.<\/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\/Ex5.2\/image005.png\" \/>, to find n<sup>th<\/sup> term of arithmetic progression,<\/p>\n<p style=\"text-align: justify;\">20.75 = 5 + (<em>n <\/em>\u2212 1) (1.75)<\/p>\n<p style=\"text-align: justify;\">\u21d2 20.75 = 5 + 1.75<em>n <\/em>\u2212 1.75<\/p>\n<p style=\"text-align: justify;\">\u21d2 17.5 = 1.75<em>n<\/em>\u21d2 <em>n <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">It means in the 10<sup>th<\/sup> week her savings become Rs 20.75.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_52\"><\/span>NCERT Solutions for Class 10 Maths Exercise 5.2<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 5.2 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 5.2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-5-2\/\" aria-label=\"More on NCERT Solutions for Class 10 Maths Exercise 5.2\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":30008,"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-4952","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 5.2 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 10 Maths Exercise 5.2 in PDF format for free download. 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