Question

The sum of an infinite GP is 57 and the sum of their cubes is 9747, find the GP.

Answer 1

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

Then, $S_{\infty }=57-\frac{a}{1-r}...\left(i\right)$

and sum of cubes = $a^3+a^3{r}^3+a^3{r}^6+a^3{r}^9+...\infty$

$\Rightarrow 9747=\frac{a^3}{1-r^3}...\left(ii\right)$

On dividing the cube of Eq. (i) by Eq. (ii), we get

$\frac{a^3}{(1-r{)}^3}\times \frac{1-r^3}{a^3}=\frac{(57{)}^3}{9747}$

$\Rightarrow \frac{1-r^3}{(1-r{)}^3}=19$

$\Rightarrow \frac{(1-r)(1+r+r^2)}{(1-r{)}^3}=19$

$\Rightarrow \frac{1+r+r^2}{(1-r{)}^2}=19$

$\Rightarrow 1+r+r^2=19(1-2r+r^2)$

$\Rightarrow 1+r+r^2=19-38r+19r^2$

$\Rightarrow 18r^2-39r+18=0$

$\Rightarrow (3r-2)(6r-9)=0$

$\Rightarrow r=\frac{2}{3}$ or $r=\frac{3}{2}$

$\therefore r=\frac{2}{3}$ [$\because r\ne 3/2$ because in infinite GP, $\left|r\right|<1$]

Substituting the value of r in Eq. (i), we get

$\frac{a}{1-2/3}=57\Rightarrow a=19$

Hence, the GP is 19,$\frac{38}{3},\frac{76}{9},....$