JEE Main 2024 — Sequences and Series Question with Solution

$\begin{aligned} & \sum_{n=0}^{\infty} \operatorname{ar}^{\mathrm{n}}=57 \\ & \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 \\ & \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots(I)\\ & \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 \\ & \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots . \ldots=9746 \\ & \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots(II)\\ & \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\frac{\mathrm{a}^3}{(1-\mathrm{r})^3}}{\frac{\mathrm{a}^3}{1-\mathrm{r}^3}}=\frac{57^3}{9717}=19 \\ & \mathrm{a}=19 \\ & \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31 \\ & \end{aligned}$<br/>Onsolving,$\mathrm{r}=\frac{2}{3}$and$\mathrm{r}=\frac{3}{2}$(rejected$)$ $\begin{array}{rl}& a=19\\ & \therefore a+18r=19+18\times \frac{2}{3}=31\end{array}$