{"id":4609,"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-9-3-part-2\/"},"modified":"2016-05-17T11:09:28","modified_gmt":"2016-05-17T05:39:28","slug":"ncert-solutions-class-11-maths-exercise-9-3-part-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-9-3-part-2\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 9.3 Part-2"},"content":{"rendered":"

16. Find a G.P. for which sum of the first two terms is and the fifth term is 4 times the third term.<\/strong><\/p>\n

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

Given:<\/p>\n

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

And<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

Therefore, required G.P. is <\/p>\n

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

<\/p>\n

Therefore, required G.P. is <\/p>\n

\n

17. If the 4th<\/sup>, 10th<\/sup> and 16th<\/sup> terms of a G.P. are and respectively. Prove that are in G.P.<\/strong><\/p>\n

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

<\/p>\n

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

<\/p>\n

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

<\/p>\n

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

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

<\/p>\n

<\/p>\n

[From eq. (i) and (iii)]\n

are in G.P.<\/p>\n

\n

18. Find the sum to terms of the sequences 8, 88, 888, 8888, \u2026\u2026<\/strong><\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

= <\/p>\n

\n

19. Find the sum of the product of the corresponding terms of the sequences 2, 4, 16, 32 and 128, 32, 8, 2, <\/strong><\/p>\n

Ans. <\/strong>Multiplying the corresponding terms of the given sequences 2, 4, 16, 32 and 128, 32, 8, 2, <\/p>\n

<\/p>\n

are in G.P.<\/p>\n

Here and <\/p>\n

when <\/p>\n

= <\/p>\n

= 496<\/p>\n

\n

20. Show that the products of the corresponding terms of the sequences and form a G.P. and find the common ratio.<\/strong><\/p>\n

Ans. <\/strong>Multiplying the corresponding terms of the given sequences, we have<\/p>\n

<\/p>\n

are in G.P.<\/p>\n

Here First term = and common ratio = <\/p>\n

\n

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than by 4th<\/sup> by 18.<\/strong><\/p>\n

Ans. <\/strong>Let the four numbers in G.P. be <\/p>\n

and <\/p>\n

Now,<\/p>\n

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

And<\/p>\n

<\/p>\n

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

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

<\/p>\n

<\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the required numbers are <\/p>\n

\n

22. If the and terms of a G.P. are and respectively. Prove that <\/strong><\/p>\n

Ans. <\/strong>Let A be the first term and R be the common ratio of given G.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, L.H.S. = <\/p>\n

= <\/p>\n

= <\/p>\n

= = = 1 = R.H.S.<\/p>\n

\n

23. If the first and the term of a G.P. are and respectively and if P is the product of terms, prove that <\/strong><\/p>\n

Ans. <\/strong>Here, first term of G.P. is <\/p>\n

And<\/p>\n

<\/p>\n

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

Given: P = <\/p>\n

P = <\/p>\n

[Squaring both sides]\n

<\/p>\n

[From eq. (i)]\n

<\/p>\n

<\/p>\n

\n

24. Show that the ratio of the sum of first terms of a G.P. to the sum of terms from to term is <\/strong><\/p>\n

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

Then<\/p>\n

= <\/p>\n

\n

25. If and are in G.P., show that <\/strong><\/p>\n

Ans. <\/strong>Let be the common ratio of given G.P.<\/p>\n

Then and <\/p>\n

Now, L.H.S. = <\/p>\n

= <\/p>\n

= = <\/p>\n

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

= <\/p>\n

= <\/p>\n

= = <\/p>\n

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

\n

26. Insert two numbers between 3 and 81 so that the resulting sequence us G.P.<\/strong><\/p>\n

Ans. <\/strong>Let be two numbers between 3 and 81 such that are in G.P.<\/p>\n

Here and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

And <\/p>\n

Therefore, the required numbers are 9 and 27.<\/p>\n

\n

27. Find the value of so that may be the geometric mean between and .<\/strong><\/p>\n

Ans. <\/strong>Since, G.M. between two numbers and is <\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio <\/strong><\/p>\n

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

Applying component and dividend, we get<\/p>\n

<\/p>\n

<\/p>\n

Again, Applying component and dividend, we get<\/p>\n

<\/p>\n

<\/p>\n

Squaring both sides,<\/p>\n

<\/p>\n

Therefore, the numbers are in the ratio <\/p>\n

\n

29. If A and G be A.M. and G.M. respectively between two positive numbers, prove that the numbers are <\/strong><\/p>\n

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

Now,<\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= = <\/p>\n

= and<\/p>\n

= and<\/p>\n

= and<\/p>\n

\n

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd<\/sup> hour, 4th<\/sup> hour and hour?<\/strong><\/p>\n

Ans. <\/strong>Bacteria present in the culture originally = 30<\/p>\n

Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.<\/p>\n

Here and <\/p>\n

Bacteria at the end of 2nd<\/sup> hour = <\/p>\n

And Bacteria at the end of 4th<\/sup> hour = <\/p>\n

And Bacteria at the end of hour = <\/p>\n

\n

31. What will Rs. 500 amount to 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?<\/strong><\/p>\n

Ans. <\/strong>Original amount = Rs. 500, Rate of interest = 10% compounded annually<\/p>\n

Interest of one year = = Rs. 50<\/p>\n

And Amount after one year = 500 + 50 = Rs. 550<\/p>\n

Here and <\/p>\n

Therefore, amount after 10 years = <\/p>\n

\n

32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively then obtain the quadratic equation.<\/strong><\/p>\n

Ans. <\/strong>Let and be the roots of required quadratic equation.<\/p>\n

Then A.M. = <\/p>\n

<\/p>\n

And G.M. = <\/p>\n

<\/p>\n

Now, Quadratic equation <\/p>\n

<\/p>\n

<\/p>\n

Therefore, required equation is .<\/p>\n","protected":false},"excerpt":{"rendered":"

16. Find a G.P. for which sum of the first two terms is and the fifth term is 4 times the third term. Ans. Let be the first term and be the common ratio of given G.P. Given: \u2026\u2026..(i) And Putting in eq. (i), we get Therefore, required G.P. is Putting in eq. (i), we … 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-4609","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 9.3 Part-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-9-3-part-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 9.3 Part-2 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"16. Find a G.P. for which sum of the first two terms is and the fifth term is 4 times the third term. Ans. Let be the first term and be the common ratio of given G.P. Given: \u2026\u2026..(i) And Putting in eq. (i), we get Therefore, required G.P. is Putting in eq. (i), we ... 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