Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

 

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

Selina Class 10 ICSE solution Chapter 11 Geometric Progression will give strength to you and the board will also clear the exam, Mathematics Subject demands time and practice, then keep practicing and you will definitely get success.

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Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

Exercise 11(B) Page No: 154

Question 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

tn = arn – 1

So,

tn = (-10)( 1/(-2√3))n – 1 = -5/72

Now, equating the exponents we have

n – 1 = 4

n = 5

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

Question 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

tn = arn – 1

So,

t5 = ar5 – 1 = ar4 = 81 …. (1)

And,

t2 = ar2 – 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, ……

Question 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

tn = arn – 1

So,

t4 = ar4 – 1 = ar3 = 1/18 …. (1)

And,

t7 = ar7 – 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, …..

Question 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

tn = arn – 1

So,

t1 = ar1 – 1 = a = 2 …. (1)

And,

t3 = ar3 – 1 = ar2 = 8 …. (2)

Dividing (2) by (1), we have

ar2/ a = 8/ 2

r2 = 4

r = ± 2

Hence, the 2nd 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

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

Solution:

Given,

Product of 3rd and 8th 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,

tn = arn – 1

So,

t3 x t8 = ar3 – 1 x ar8 – 1 = ar2 x ar7 = a2r9 = 243

Also given,

t4 = ar4 – 1 = ar3 = 3

Now,

a2r9 = (ar3) ar6 = 243

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

(3) ar6 = 243

ar6 = 81

ar7 – 1 = 81 = t7

Thus, the 7th term of the G.P is 81.

Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

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