Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& \sum _{n=0}^{\infty }{\mathrm{ar}}^{\mathrm{n}}=57\\ & \mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2+\infty =57\\ & \frac{\mathrm{a}}{1-\mathrm{r}}=57\dots (I)\\ & \sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}^3{\mathrm{r}}^{3\mathrm{n}}=9747\\ & {\mathrm{a}}^3+{\mathrm{a}}^3\cdot {\mathrm{r}}^3+{\mathrm{a}}^3\cdot {\mathrm{r}}^6+\dots \dots \dots .\dots =9746\\ & \frac{{\mathrm{a}}^3}{1-{\mathrm{r}}^3}=9746\dots (II)\\ & \frac{(\mathrm{I}{)}^3}{(\mathrm{II})}\Rightarrow \frac{\frac{{\mathrm{a}}^3}{(1-\mathrm{r}{)}^3}}{\frac{{\mathrm{a}}^3}{1-{\mathrm{r}}^3}}=\frac{57^3}{9717}=19\\ & \mathrm{a}=19\\ & \therefore \mathrm{a}+18\mathrm{r}=19+18\times \frac{2}{3}=31\\ & \end{array}$
On solving, $\mathrm{r}=\frac{2}{3}$ and $\mathrm{r}=\frac{3}{2}$ (rejected $)$ $\begin{array}{rl}& a=19\\ & \therefore a+18r=19+18\times \frac{2}{3}=31\end{array}$