JEE Main 2025 — Mathematics Algebra

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

$\begin{aligned}
\mathrm{a}_3 \mathrm{a}_5 & =\operatorname{ar}^2 \cdot \operatorname{ar}^4=729 \
& =\mathrm{a}^2 \mathrm{r}^6=729 \
& =\mathrm{ar}^3=27 \quad ....(i)
\end{aligned}$

$$\begin{aligned} \mathrm{a}_2+\mathrm{a}_4 & =\mathrm{ar}+\mathrm{ar}^3=\frac{111}{4} \\ & =\mathrm{ar}=\frac{3}{4} \quad ....(ii)\end{aligned}$$a2​+a4​​=ar+ar3=4111​=ar=43​....(ii)​

$$\text { (i) } \div \text { (ii) }$$ (i) ÷ (ii) 

$\begin{aligned}
& \frac{\mathrm{ar}^3}{\mathrm{ar}}=\frac{27}{3 / 4} \
& \mathrm{r}^2=36 \
& \mathrm{r}=6
\end{aligned}$

from (ii)

$$a(6)=\frac{3}{4} \Rightarrow a=\frac{1}{8}$$a(6)=43​⇒a=81​

Now, $24\left(a_1+a_2+a_3\right)$24(a1​+a2​+a3​)

$\begin{aligned}
& =24\left(a+a r+a^2\right) \
& =24 a\left(1+r+r^2\right)
\end{aligned}$

$$\begin{aligned} & =24 \times \frac{1}{8}(1+6+36) \\ & =3(43) \\ & =129\end{aligned}$$​=24×81​(1+6+36)=3(43)=129​