The sum of an infinite G.P. is 57 and the sum of their cubes is 9747, then the common ratio of the G.P. is

Let a be the first term and r the common ratio of the G.P. Then, the sum is given by

$\frac{\mathrm{a}}{1-\mathrm{r}}=57$              (1)

Sum of the cubes is 9747. Hence,

$\begin{array}{l}{\mathrm{a}}^3+{\mathrm{a}}^3{\mathrm{r}}^3+{\mathrm{a}}^3{\mathrm{r}}^6+\cdots =9747\\ \Rightarrow \frac{{\mathrm{a}}^3}{1-{\mathrm{r}}^3}=9747 \left(2\right)\end{array}$

Dividing the cube of (1) by (2), we get

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

$\begin{array}{l}\mathrm{or} \frac{1-{\mathrm{r}}^3}{(1-\mathrm{r}{)}^3}=19\\ \mathrm{or} \frac{1+\mathrm{r}+{\mathrm{r}}^2}{(1-\mathrm{r}{)}^2}=19\\ \mathrm{or} 18{\mathrm{r}}^2-39\mathrm{r}+18=0\\ \mathrm{or} (3\mathrm{r}-2)(6\mathrm{r}-9)=0\end{array}$

$\mathrm{or} \mathrm{r}=2/3[\because \mathrm{r}\ne 3/2\text{, because }0<|\mathrm{r}|<1\text{ for an infinite G.P.] }$