Question

The sum of an infinite number of terms of a G.P. is $4$, and the sum of their cubes is $192$; find the series.

Answer 1

Let $S$ be the sum of infinite terms of G.P. having first term $=a$ and common ratio $=r$
$S=4$ ..... (given)
$S=\frac{a}{1-r}=4$
Cubing each term of G.P., first term $=a^3$ and common ratio $=r^3$
Now the sum is

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

$\Rightarrow \frac{a.a^2}{(1-r)(r^2+r+1)}=192$
$\Rightarrow \frac{4a^2}{(r^2+r+1)}=192$

$\Rightarrow \frac{a^2}{(r^2+r+1)}=48$
$\Rightarrow \frac{16(1-r{)}^2}{(r^2+r+1)}=48$
$\Rightarrow (r^2-2r+1)=3(r^2+r+1)$
$\Rightarrow 2r^2+5r+2=0$
$\Rightarrow r=-2$ or $\frac{-1}{2}$
$r=-2$ will be discarded as above formulae is valid only for $\left|r\right|<1$
$\therefore r=\frac{-1}{2}$
$\Rightarrow a=6$
Thus, the series is $6,-3,\frac{3}{2},......\infty$