Question

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

• A. $1/2$
• B. $2/3$
• C. $1/3$
• D. $-1/2$

Answer 1

The correct option is D $-1/2$

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

$s_{\infty }=\frac{a}{1-r}$

$\Rightarrow 4=\frac{a}{1-r}$

$\Rightarrow a=4(1-r)$ ……..$\left(1\right)$

Now, when the terms are cubed

First term$=a^3$ and common ratio$=r^3$

$\frac{a^3}{1-r^3}=192$

$\Rightarrow a^3=192(1-r^3)$ …………..$\left(2\right)$

From equation $\left(1\right)$

$a=4(1-r)$

Cubing both sides

$a^3=64(1-r{)}^3$

Put value of $a^3$ in equation $\left(2\right)$

$64(1-r{)}^3=192(1-r\left)\right(1+r^2+r)$

$(1-r{)}^2=3(1+r^2+r)$

$1+r^2-2r=3+3r^2+3r$

$2r^2+5r+2=0$

By splitting middle term

$2r^2+4r+r+2=0$

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

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

$\Rightarrow r=\frac{-1}{2}$.