Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Exercise 11(B)

Problems on finding the G.P. when certain terms or conditions are given are included in this exercise. For any clarifications on how to solve problems of this chapter and others, students can refer to the Selina Solutions for Class 10 Maths anytime. It is a vital resource for exam preparations. Further, the solutions to this exercise can be accessed in the Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Exercise 11(B) PDF, which is given in the links below.

Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Exercise 11(B) Download PDF

Access Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Exercise 11(B)

Exercise 11(B) Page No: 154

1. Which term of the G.P. :

Solution:

In the given G.P.

First term, a = -10

Common ratio, r = (5/√3)/ (-10) = 1/(-2√3)

We know that, the general term is

t_n = ar^{n – 1}

So,

t_n = (-10)( 1/(-2√3))^{n – 1} = -5/72

Now, equating the exponents we have

n – 1 = 4

n = 5

Thus, the 5^th of the given G.P. is -5/72

2. The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

Solution:

Given,

t₅ = 81 and t₂ = 24

We know that, the general term is

t_n = ar^{n – 1}

So,

t₅ = ar^{5 – 1} = ar⁴ = 81 …. (1)

And,

t₂ = ar^{2 – 1} = ar¹ = 24 …. (2)

Dividing (1) by (2), we have

ar⁴/ ar = 81/ 24

r³ = 27/ 8

r = 3/2

Using r in (2), we get

a(3/2) = 24

a = 16

Hence, the G.P. is

G.P. = a, ar, ar², ar³……

= 16, 16 x (3/2), 16 x (3/2)², 16 x (3/2)³

= 16, 24, 36, 54, ……

3. Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P.

Solution:

Given,

t₄ = 1/18 and t₇ = -1/486

We know that, the general term is

t_n = ar^{n – 1}

So,

t₄ = ar^{4 – 1} = ar³ = 1/18 …. (1)

And,

t₇ = ar^{7 – 1} = ar⁶ = -1/486 …. (2)

Dividing (2) by (1), we have

ar⁶/ ar³ = (-1/486)/ (1/18)

r³ = -1/27

r = -1/3

Using r in (1), we get

a(-1/3)³ = 1/18

a = -27/ 18 = -3/2

Hence, the G.P. is

G.P. = a, ar, ar², ar³……

= -3/2, -3/2(-1/3), -3/2(-1/3)², -3/2(-1/3)³, ……

= -3/2, 1/2, -1/6, 1/18, …..

4. If the first and the third terms of a G.P are 2 and 8 respectively, find its second term.

Solution:

Given,

t₁ = 2 and t₃ = 8

We know that, the general term is

t_n = ar^{n – 1}

So,

t₁ = ar^{1 – 1} = a = 2 …. (1)

And,

t₃ = ar^{3 – 1} = ar² = 8 …. (2)

Dividing (2) by (1), we have

ar²/ a = 8/ 2

r² = 4

r = ± 2

Hence, the 2^nd term of the G.P. is

When a = 2 and r = 2 is 2(2) = 4

Or when a = 2 and r = -2 is 2(-2) = -4

5. The product of 3^rd and 8^th terms of a G.P. is 243. If its 4^th term is 3, find its 7^th term

Solution:

Given,

Product of 3^rd and 8^th terms of a G.P. is 243

The general term of a G.P. with first term a and common ratio r is given by,

t_n = ar^{n – 1}

So,

t₃ x t₈ = ar^{3 – 1} x ar^{8 – 1} = ar² x ar⁷ = a²r⁹ = 243

Also given,

t₄ = ar^{4 – 1} = ar³ = 3

Now,

a²r⁹ = (ar³) ar⁶ = 243

Substituting the value of ar³in the above equation, we get,

(3) ar⁶ = 243

ar⁶ = 81

ar^{7 – 1} = 81 = t₇

Thus, the 7^th term of the G.P is 81.

Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(D) Solutions