Question

The sum of nfinite number of terms of a G.P. is 4 and the sum of their cubes is 192 . Find the series.

Answer 1

Let $a,ar,a{r}^2,a{r}^3,...$ be a G.P, where $a$ is the initial term and $r$ is the common ratio.

Given:Sum of infinite terms of G.P$=4$

Sum of infinite terms of G.P$=\frac{a}{1-r}$

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

Cubing the above equation, we have

$\frac{a^3}{{(1-r)}^3}=64$ ......$\left(2\right)$

Now, consider the cubes of the above terms:

Thus,$a^3,a^3{r}^3,a^3{r}^6,a^3{r}^9,....$

$A=a^3$ and $R=r^3$

Sum of infinite terms$=\frac{A}{1-R}=192$

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

Dividing equation $\left(3\right)$ by equation $\left(2\right),$ we have,

$\frac{a^3}{1-r^3}\times \frac{{(1-r)}^3}{a^3}=\frac{192}{64}$

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

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

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

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

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

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

$\Rightarrow r+2=0,2r+1=0$

Since $\left|r\right|<1,$ we have $r\ne -2$

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

From $\left(1\right)\frac{a}{1-r}=4$

$\Rightarrow \frac{a}{1+\frac{1}{2}}=4$

$\Rightarrow \frac{2a}{3}=4$

$\therefore a=6$

Hence $a=6,r=-\frac{1}{2}$

Hence the series is $6,6\times \frac{-1}{2},6\times \frac{1}{4},6\times \frac{-1}{8},...$

or $6,-3,\frac{3}{2},\frac{-3}{4},...$