{"id":4611,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-1\/"},"modified":"2016-05-17T11:13:17","modified_gmt":"2016-05-17T05:43:17","slug":"ncert-solutions-class-11-maths-miscellaneous-part-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-1\/","title":{"rendered":"NCERT Solutions class-11 Maths Miscellaneous Part-1"},"content":{"rendered":"

Miscellaneous Exercise<\/strong><\/p>\n

1. Show that the sum of <\/strong>\u00a0and <\/strong>\u00a0terms of an A.P. is equal to twice the <\/strong>\u00a0terms.<\/strong><\/p>\n

Ans. <\/strong>Here, \u00a0\u2026..(i)<\/p>\n

and \u00a0\u00a0\u2026..(ii)<\/p>\n

To prove: <\/p>\n

Adding eq. (i) and (ii), we get<\/p>\n

\u00a0 <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\n

2. If the sum of three numbers in A.P., is 24 and their product is 440. Find the numbers.<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0be three numbers in A.P.<\/p>\n

According to question,\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

And <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Taking , A.P. is (8 \u2013 3), 8, (8 + 3)<\/p>\n

\u00a05, 8, 11<\/p>\n

Taking \u00a0A.P. is (8 + 3), 8, (8 \u2013 3)<\/p>\n

\u00a011, 8, 5<\/p>\n

\n

3. Let sum of <\/strong>\u00a0terms of an A.P. be respectively, show that.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0\u00a0\u2026..(i)<\/p>\n

\u2026..(ii)<\/p>\n

And <\/p>\n

Now, <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0 Proved.<\/p>\n

\n

4. Find the sum of all numbers between 200 and 400 which are divisible by 7.<\/strong><\/p>\n

Ans. <\/strong>Given: A.P. 203, 210, 217, \u2026\u2026\u2026., 399<\/p>\n

Here \u00a0and <\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 8729<\/p>\n

\n

5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.<\/strong><\/p>\n

Ans. <\/strong>Given: A.P. which is divisible by 2<\/p>\n

2, 4, 6, \u2026\u2026\u2026., 100<\/p>\n

Here \u00a0and <\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 2550<\/p>\n

Again A.P. which is divisible by 5<\/p>\n

5, 10, 15, \u2026\u2026\u2026., 100<\/p>\n

Here \u00a0and <\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 1050<\/p>\n

Again A.P. which is divisible by both 2 and 5<\/p>\n

10, 20, 30, \u2026\u2026\u2026., 100<\/p>\n

Here \u00a0and <\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 550<\/p>\n

Now, According to question, \u00a0= (2550 + 1050) \u2013 550 = 3050<\/p>\n

\n

6. Find the sum of all two digit numbers which when divided by 4, yield 1 as remainder.<\/strong><\/p>\n

Ans. <\/strong>Given: A.P. 13, 17, 21, \u2026\u2026\u2026., 97<\/p>\n

Here \u00a0and <\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 1210<\/p>\n

\n

7. If <\/strong>\u00a0is a function satisfying <\/strong>\u00a0for all <\/strong>\u00a0N such that <\/strong>\u00a0=3 and <\/strong>\u00a0= 120, find the value of <\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Here: \u00a0and \u00a0for all \u00a0N \u2026\u2026\u2026.(i)<\/p>\n

Putting \u00a0in eq. (i), <\/p>\n

\u00a0\u00a0\u00a0= 9<\/p>\n

Putting \u00a0in eq. (i), <\/p>\n

\u00a0\u00a0\u00a0\u00a0= 27<\/p>\n

Putting \u00a0in eq. (i), <\/p>\n

\u00a0\u00a0\u00a0= 81<\/p>\n

Now, <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0and <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\n

9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find common ratio of G.P.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0and <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

= \u00a0= <\/p>\n

= <\/p>\n

\u00a0or \u00a0which is not possible<\/p>\n

Therefore, common ratio is <\/p>\n

\n

10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0be three numbers in G.P., therefore, <\/p>\n

\u00a0\u00a0 ..(i)<\/p>\n

According to question,\u00a0 \u00a0are in A.P.<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u2026..(ii)<\/p>\n

Dividing eq. (i) by eq. (ii), <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

= \u00a0= \u00a0= <\/p>\n

\u00a0or <\/p>\n

Putting \u00a0in eq. (i),\u00a0 <\/p>\n

\u00a0<\/p>\n

Then the required numbers are 8, 16, 32.<\/p>\n

Putting \u00a0in eq. (i),\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Then the required numbers are 32, 16, 8.<\/p>\n

\n

11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.<\/strong><\/p>\n

Ans. <\/strong>Let the number of terms be , number of odd terms be \u00a0and \u00a0be in G.P.<\/p>\n

\u00a0\u00a0\u00a0\u00a0and <\/p>\n

According to question,\u00a0 <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0and <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Also, <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

13. If <\/strong>\u00a0then show that <\/strong>\u00a0and <\/strong>\u00a0are in G.P.<\/strong><\/p>\n

Ans. <\/strong>Taking <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Taking <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

From eq. (i) and (ii),\u00a0 <\/p>\n

\n

14. Let S be the sum, P the product and R the sum of reciprocals of <\/strong>\u00a0terms in a G.P. Prove that <\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Here , P = \u00a0and R = <\/p>\n

Now <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Proved.<\/p>\n

\n

15. The <\/strong>\u00a0and <\/strong>\u00a0terms of an A.P. are <\/strong>\u00a0respectively. Show that <\/strong><\/p>\n

\u00a0<\/strong><\/p>\n

Ans. <\/strong>According to question,\u00a0 , \u00a0and <\/p>\n

Now <\/p>\n

Putting values of \u00a0and \u00a0we get<\/p>\n

\u00a0 <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

<\/p>\n

\u00a00 = 0 Proved.<\/p>\n

\n

16. If <\/strong>\u00a0are in A.P., prove that <\/strong>\u00a0are in A.P.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0are in A.P.<\/p>\n

\u00a0\u00a0\u00a0are in A.P.<\/p>\n

\u00a0\u00a0\u00a0are in A.P.<\/p>\n

\u00a0\u00a0\u00a0are in A.P. [Adding 1 to each numerator]\n

\u00a0\u00a0\u00a0are in A.P.<\/p>\n

\u00a0\u00a0\u00a0are in A.P.[Dividing each fraction by ]\n

\u00a0\u00a0\u00a0are in A.P.[Multiplying each fraction by ]\n

\u00a0\u00a0\u00a0are in A.P.<\/p>\n","protected":false},"excerpt":{"rendered":"

Miscellaneous Exercise 1. Show that the sum of \u00a0and \u00a0terms of an A.P. is equal to twice the \u00a0terms. Ans. Here, \u00a0\u2026..(i) and \u00a0\u00a0\u2026..(ii) To prove: Adding eq. (i) and (ii), we get \u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 2. If the sum of three numbers in A.P., is 24 and their product is 440. … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47,281],"tags":[],"class_list":["post-4611","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Miscellaneous Part-1 | myCBSEguide<\/title>\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-11-maths-miscellaneous-part-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions class-11 Maths Miscellaneous Part-1 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Miscellaneous Exercise 1. Show that the sum of \u00a0and \u00a0terms of an A.P. is equal to twice the \u00a0terms. Ans. Here, \u00a0\u2026..(i) and \u00a0\u00a0\u2026..(ii) To prove: Adding eq. (i) and (ii), we get \u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 2. If the sum of three numbers in A.P., is 24 and their product is 440. ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-1\/\" \/>\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=\"2016-05-17T04:19:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2016-05-17T05:43:17+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/11\/maths\/ch09\/Misc\/image001.png\" \/>\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 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