Question

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 192. The common ratio of the original G.P. is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D

$-\frac{1}{2}$

(d) $-\frac{1}{2}$

Let a be the first term and r be the common ratio of the given G.P. then,

Sum = 4

$\Rightarrow \frac{a}{1-r}=4$ ....... (i)

Sum of cubes = 192

$\Rightarrow a^3+a^3{r}^3+a^3{r}^6+....=192$

$\Rightarrow \frac{a^3}{1-r^3}=192$ ....... (ii)

Dividing the cube of (i) by (ii) we get

$\frac{a^3}{1-r^3}.\frac{1-r^3}{a^3}=\frac{(4{)}^3}{192}$

$\Rightarrow \frac{1-r^3}{(1-r{)}^3}=\frac{1}{3}$

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

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

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

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

since $r\ne -2$ because - 1 < r < 1 for an infinite G.P.

Thus, $r=\frac{-1}{2}$