{"id":4607,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-9-2\/"},"modified":"2016-05-17T11:07:20","modified_gmt":"2016-05-17T05:37:20","slug":"ncert-solutions-class-11-maths-exercise-9-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-9-2\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 9.2"},"content":{"rendered":"

Exercise 9.2<\/strong><\/p>\n

1. Find the sum of odd integers from 1 to 2001.<\/strong><\/p>\n

Ans. <\/strong>Odd integers from 1 to 2001 are 1, 3, 5, 7, \u2026\u2026.., 2001.<\/p>\n

Here, and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now,<\/p>\n

<\/p>\n

= 1002001<\/p>\n

\n

2. Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.<\/strong><\/p>\n

Ans. <\/strong>According to question, series is105, 110, 115, 120, \u2026\u2026\u2026, 995<\/p>\n

Here and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now,<\/p>\n

<\/p>\n

= 98450<\/p>\n

\n

3. In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th<\/sup> term is <\/strong><\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

4. How many terms of the A.P., <\/strong> are needed to give the sum <\/strong> ?<\/strong><\/p>\n

Ans. <\/strong>Here,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

or <\/p>\n

\n

5. In an A.P., if <\/strong> term is <\/strong> and <\/strong> term is <\/strong> prove that the sum of first <\/strong> terms is <\/strong> where <\/strong><\/p>\n

Ans. <\/strong>Let be the first term and be the common difference of given A.P.<\/p>\n

And and <\/p>\n

and<\/p>\n

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

Subtracting eq. (ii) from eq. (i), we get<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Putting value of in eq. (i), we get<\/p>\n

<\/p>\n

<\/p>\n

Now,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

<\/p>\n

\n

6. If the sum of a certain number of terms of the A.P. 25, 22, 19, \u2026.. is 116, find the last term.<\/strong><\/p>\n

Ans. <\/strong>Here and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

or<\/p>\n

or<\/p>\n

But is not possible. Therefore, <\/p>\n

Now,<\/p>\n

<\/p>\n

<\/p>\n

\n

7. Find the sum of <\/strong> terms of an A.P. whose <\/strong> term is <\/strong><\/p>\n

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

Putting and <\/p>\n

and <\/p>\n

<\/p>\n

<\/p>\n

\n

8. If the sum of <\/strong> terms of an A.P. is <\/strong> where <\/strong> and <\/strong> are constants, find the common difference.<\/strong><\/p>\n

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

Replacing by <\/p>\n

<\/p>\n

= <\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

Again here replacing by <\/p>\n

<\/p>\n

= <\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

\n

9. The sums of <\/strong> terms of two arithmetic progressions are on the ratio <\/strong> Find the ratio of their 18th<\/sup> terms.<\/strong><\/p>\n

Ans. <\/strong>Let and be the first terms and common differences of two A.P\u2019s respectively.<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now, to get 18th<\/sup> term,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the ratio of 18th<\/sup> terms of two A.P.\u2019s is 179: 321.<\/p>\n

\n

10. If the sum of first <\/strong> terms of an A.P. is equal to the sum of the first <\/strong> terms, then find the sum of the first <\/strong> terms.<\/strong><\/p>\n

Ans. <\/strong>Let be the first term and be the common difference of given A.P.<\/p>\n

and <\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now<\/p>\n

= <\/p>\n

<\/p>\n

\n

11. Sum of the first <\/strong> and <\/strong> terms of an A.P. are <\/strong> and <\/strong> respectively. Prove that <\/strong><\/p>\n

Ans. <\/strong>Let A be the first term and be the common difference of given A.P.<\/p>\n

<\/p>\n

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

<\/p>\n

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

<\/p>\n

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

Now<\/p>\n

Putting the values of and from eq. (i), (ii) and (iii), we get<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

L.H.S. = R.H.S. Proved.<\/p>\n

\n

12. The ratio of the sum of <\/strong> and <\/strong> terms of an A.P. is <\/strong> Show that the ratio of <\/strong> and <\/strong> term is <\/strong><\/p>\n

Ans. <\/strong>Let be the first term and be the common difference of given A.P.<\/p>\n

and <\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now, = <\/p>\n

= <\/p>\n

<\/p>\n

\n

13. If the sum of <\/strong> terms of an A.P. is <\/strong> and its <\/strong> term is 164, find the value of <\/strong><\/p>\n

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

Replacing by in we get<\/p>\n

<\/p>\n

<\/p>\n

= = <\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

And<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

14. Insert five numbers between 8 and 26 so that the resulting sequence is an A.P.<\/strong><\/p>\n

Ans. <\/strong>Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub> and A5<\/sub> be five numbers between 8 and 26.<\/p>\n

8, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, 26<\/p>\n

Here, and and let be the common difference.<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now, A1<\/sub> = <\/p>\n

A2<\/sub> = <\/p>\n

A3<\/sub> = <\/p>\n

A4<\/sub> = <\/p>\n

A5<\/sub> = <\/p>\n

\n

15. If <\/strong> is the A.M. between <\/strong> and <\/strong> then find the value of <\/strong><\/p>\n

Ans. <\/strong>Since, A.M. between and is <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

16. Between 1 and 31, <\/strong> numbers have been inserted in such a way that resulting sequence is an A.P. and the ratio of 7th<\/sup> and <\/strong> numbers is 5: 9. Find the value of <\/strong><\/p>\n

Ans. <\/strong>Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, \u2026\u2026.., Am<\/sub> be numbers between 1 and 31.<\/p>\n

Here, and let the common difference be <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now,<\/p>\n

And<\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

17. A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th<\/sup> installment?<\/strong><\/p>\n

Ans. <\/strong>Amount of 1st<\/sup> installment = Rs. 100 and Amount of 2nd<\/sup> installment = Rs. 105<\/p>\n

and <\/p>\n

Now<\/p>\n

<\/p>\n

<\/p>\n

= Rs. 245<\/p>\n

Therefore, the amount of 30th<\/sup> installment is Rs. 245.<\/p>\n

\n

18. The difference between any two consecutive interior angles of a polygon is <\/strong> If the smallest angle is <\/strong> find the number of the sides of the polygon.<\/strong><\/p>\n

Ans. <\/strong>Let the number of sides of polygon be The interior angles of the polygon form an A.P.<\/p>\n

Here, and <\/p>\n

Since Sum of interior angles of a polygon with sides is <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

or <\/p>\n

But not possible because == > <\/p>\n

Therefore, number of sides of the polygon are 9.<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise 9.2 1. Find the sum of odd integers from 1 to 2001. Ans. Odd integers from 1 to 2001 are 1, 3, 5, 7, \u2026\u2026.., 2001. Here, and Now, = 1002001 2. Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5. Ans. According to question, series … 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-4607","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 9.2 | 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-exercise-9-2\/\" \/>\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 Exercise 9.2 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exercise 9.2 1. Find the sum of odd integers from 1 to 2001. Ans. Odd integers from 1 to 2001 are 1, 3, 5, 7, \u2026\u2026.., 2001. Here, and Now, = 1002001 2. Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5. Ans. According to question, series ... 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