JEE Main 2024 — 9 April, Shift 2 — Question 16

$\quad \sum_{\mathrm{n}=0}^{\infty} \mathrm{ar}^{\mathrm{n}}=57$

$a+a r+a r^{2}+ \ldots \infty=57$

$\frac{a}{1-r}=57$

$\sum_{n=0}^{\infty} a^{3} r^{3 n}=9747$

$a^{3}+a^{3} \cdot r^{3}+a^{3} \cdot r^{6}+ \ldots \infty=9746$

$\frac{a^{3}}{1-r^{3}}=9746 \ldots \frac{(\mathrm{I})^{3}}{\text { (II) }}$

$\Rightarrow \frac{\frac{a^{3}}{(1-r)^{3}}}{\frac{a^{3}}{1-r^{3}}}=\frac{57^{3}}{9717}=19$

On solving, $r=\frac{2}{3}$ and $r=\frac{3}{2}$ (rejected) $\mathrm{a}=19$

$\therefore a+18 r=19+18 \times \frac{2}{3}=31$