Find the sum of 6 terms of the G.P. 1 , - 2 3 , 4 9 , . . . .

Given sequence is, 1 , - 2 3 , 4 9 , . . . . The sequence forms a G.P where a = 1 and n = 6 We know that, common ratio in a G . P . is r = a n a n - 1 ∴ Common ratio r = - 2 3 1 = - 2 3 < 1 We know that, sum of n terms in G.P is S n = a 1 - r n 1 - r ∴ S 6 = 1 1 - - 2 3 6 1 - - 2 3 = 1 1 - 64 729 1 + 2 3 = 729 - 64 729 3 + 2 3 = 665 729 5 3 = 665 729 × 3 5 = 243 133 . ∴ The sum of 10 terms of the given series is 243 133 .

Find the value of n , if the sum of first n terms of 2 9 - 1 3 + 1 2 . . . . is 55 72 .

Given, series is 2 9 - 1 3 + 1 2 . . . . . Here, first term, a = 2 9 , a 2 = - 1 3 , a 3 = 1 2 As, a 2 a = - 1 3 2 9 = - 3 2 and a 3 a 2 = 1 2 - 1 3 = - 3 2 so, the given series is a G.P. series, with common ratio, r = - 3 2 Also, given, sum of n terms, S n = 55 72 . ⇒ a r n - 1 r - 1 = 55 72 ⇒ 2 9 - 3 2 n - 1 - 3 2 - 1 = 55 72 ⇒ - 3 2 n - 1 = 55 72 8 × 9 2 × - 5 2 ⇒ - 3 2 n = - 275 32 + 1 ⇒ - 3 2 n = - 275 + 32 32 = - 243 32 = - 3 2 5 ∴ n = 5 .

The 2 n d and 5 t h terms of a geometric series are - 1 2 and 1 16 respectively. Find the sum of the series upto 8 terms.

let a be the first term and r be the common ratio of the given G . P . Given that, a 2 = - 1 2 We know that, n t h term of a G . P . is a n = a r n - 1 . a r = - 1 2 (i) And a 5 = 1 16 ⇒ a r 4 = 1 16 (ii) On dividing equation (ii) by equation (i), we get r 3 = 1 16 - 1 2 = - 1 8 = - 1 2 3 ⇒ r = - 1 2 ⇒ | r | = 1 2 < 1 From (i), a × - 1 2 = - 1 2 ⇒ a = 1 We know that, sum of n terms in a G . P . is S n = a 1 - r n 1 - r . ∴ S 8 = 1 1 - - 1 2 8 1 - - 1 2 = 2 3 1 - 1 256 = 2 × 255 3 × 256 = 85 128 ∴ The sum of the given series upto 8 terms is 85 128 .

The first term of a G . P . is 27 and 8 t h term is 1 81 . Find the sum of its first 10 terms.

Let a be the first term and r be the common ratio of the given G . P . Here, a = 27 and a 8 = 1 81 We know that, n t h term of a G . P . is a n = a r n - 1 . Now, a r 7 = 1 81 ⇒ 27 r 7 = 1 81 ⇒ r 7 = 1 27 × 81 = 1 3 3 × 3 4 = 1 3 7 ⇒ r = 1 3 < 1 We know that, sum of n terms in a G . P . is S n = a 1 - r n 1 - r ∴ S 10 = 27 1 - 1 3 10 1 - 1 3 = 81 2 1 - 1 3 10 ∴ The sum of its first 10 terms of the given series is 81 2 1 - 1 3 10 .

If the first term of a G . P . is 5 and the sum of first three terms is 31 5 , find the common ratio.

Let a be the first term and r be the common ratio of the given G . P . respectively. Given that, a = 5 and S 3 = 31 5 . We know that, sum of n terms of a G . P . is S n = a 1 - r n 1 - r . Now, S 3 = 5 l - r 3 1 - r ⇒ 31 5 = 5 1 - r 3 1 - r ⇒ 21 25 = 1 + r 2 + r [ ∵ r ≠ 1 ] ⇒ 25 r 2 + 25 r - 6 = 0 ⇒ 25 r 2 + 30 r - 5 r - 6 = 0 ⇒ ( 5 r + 6 ) ( 5 r - 1 ) = 0 ⇒ r = - 6 5 , 1 5 Hence, the required common ratio of G . P . is - 6 5 or 1 5 .

E M B I B E

Understanding ICSE Mathematics Class 10>Algebra>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5

M L Aggarwal Solutions for Exercise 5: Exercise 9.5

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Exercise 5: Exercise 9.5

Attempt the practice questions from Exercise 5: Exercise 9.5 with hints and solutions to strengthen your understanding. Understanding ICSE Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Exercise 5: Exercise 9.5 with Hints & Solutions

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5>Q 1

Find the sum of $6$ terms of the G.P. $1, - \frac{2}{3}, \frac{4}{9}, . . . .$

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5>Q 3

The $= \frac{[\frac{665}{729}]}{\frac{5}{3}}$ term of a G.P. is $128$ and the sum of its $n$ terms is $255$ . If its common ratio is $2$ , then find its first term.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5>Q 4

How many terms of the G.P. $1, 4, 16, . . .$ must be taken to have their sum equal to $341$ ?

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5>Q 5

Find the value of $n$ , if the sum of first $n$ terms of $\frac{2}{9} - \frac{1}{3} + \frac{1}{2} . . . .$ is $\frac{55}{72}$ .

EASY

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 9 - Arithmetic and Geometric Progressions>Exercise 9.5>Q 6

The $2^{n d}$ and $5^{t h}$ terms of a geometric series are $- \frac{1}{2}$ and $\frac{1}{16}$ respectively. Find the sum of the series upto $8$ terms.