Question

The sum of an infinite GP is 4 and the sum of the cubes of its terms is 92. The common ratio of the
original GP is

a)1/2
b)2/3
c)1/3
d)-1/2​
please answer it

Answer 1

Let the first term of infinite GP be a and common ratio be r.

s

∞

=

1−r

a

⇒4=

1−r

a

⇒a=4(1−r) ……..(1)

Now, when the terms are cubed

First term=a

3

and common ratio=r

3

1−r

3

a

3

=192

⇒a

3

=192(1−r

3

) …………..(2)

From equation (1)

a=4(1−r)

Cubing both sides

a

3

=64(1−r)

3

Put value of a

3

in equation (2)

64(1−r)

3

=192(1−r)(1+r

2

+r)

(1−r)

2

=3(1+r

2

+r)

1+r

2

−2r=3+3r

2

+3r

2r

2

+5r+2=0

By splitting middle term

2r

2

+4r+r+2=0

2r(r+2)+1(r+2)=0

(2r+1)(r+2)=0

⇒r=

2

−1

.