Question 32

The sum of an infinite geometric series of real numbers is 14, and the sum of the cubes of the terms of this series is 392. The first term of the series is

Solution

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

So, Sum of infinite GP=Â $$\ \frac{\ a}{1-r}=14$$ .............. (1)

The cubes of the terms of original GP=Â $$a^3,\ \left(ar\right)^3,\ \left(ar^2\right)^3...$$= $$a^3,\ a^3r^3,\ a^3r^6...$$

So, in the new infinite GP, first term isÂ $$a^3$$ and the common ratio isÂ $$r^3$$

Therefore, sum to infinite terms=Â $$\frac{\ a^3\ }{1-r^3}$$ ...............(2)

Cubing the eqn 1 and dividing by eqn 2, we get,

$$\frac{\ \frac{\ a^3}{\left(1-r\right)^3}\ }{\ \frac{\ a^3}{1-r^3}}=\ \ \frac{\ 14\times\ 14\times\ 14}{392}$$

=>$$\ \frac{1-r^3\ }{\left(1-r\right)^3}\ =\ \ \frac{\ 14\times\ 14\times\ 14}{392}$$

=>Â $$\ \frac{\left(1-r\right)\left(1+r^2+r\right)}{\left(1-r\right)^3}\ =\ \ \ 7$$

=>$$\ \frac{\left(1+r^2+r\right)}{\left(1-r\right)^2}\ =\ \ \ 7$$

=>Â $$\ \left(1+r^2+r\right)\ =\ \ 7\left(1+r^2-2r\right)$$

=>$$\ 1+r^2+r\ =\ \ 7+7r^2-14r$$

=>$$\ -6-6r^2+15r\ =\ 0$$

=>$$2r^2-5r+2\ =\ 0$$

=>$$2r^2-4r-r+2\ =\ 0$$

=>$$2r\left(r-2\right)-1\left(r-2\right)=0$$

=>(2r-1)(r-2)=0

.'.r= 1/2 or 2.

In v\case of infinite GP, -1<r<1.

So, r= 1/2

Putting this value in eqn 1, we get,

$$\ \frac{\ a}{1-\ \frac{\ 1}{2}}=14$$

=>2a=14

.'. a=7- Option C.

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