JEE Main 2025 — 3 April, Morning Shift — Question 35

Let the $\mathrm{I}^{\text {st }}$ term of G.P. be a $\&$ common ratio be r

$a_{3} a_{5}=a r^{2} \cdot a r^{4}=729$

$$\begin{aligned} & =\mathrm{a}^{2} \mathrm{r}^{6}=729 \\& =\mathrm{ar}^{3}=27 …(i) \end{aligned}$$

$a_{2}+a_{4}=a r+a r^{3}=\frac{111}{4}$

$$\begin{gathered} =\mathrm{ar}=\frac{3}{4} …(ii) \end{gathered}$$

(i) $\div$ (ii)

$\frac{\mathrm{ar}^{3}}{\mathrm{ar}}=\frac{27}{3 / 4}$

$r^{2}=36$ $\mathrm{r}=6$

from (ii) $\mathrm{a}(6)=\frac{3}{4} \Rightarrow \mathrm{a}=\frac{1}{8}$

Now, $24\left(a_{1}+a_{2}+a_{3}\right)$

$=24\left(a+a r+a r^{2}\right)$

$=24 \mathrm{a}\left(1+\mathrm{r}+\mathrm{r}^{2}\right)$

$=24 \times \frac{1}{8}(1+6+36)$

$=3 (43)$

$=129$