JEE Advanced 2025 — Paper 1 — Question 26

$\quad \because f(x+y)=f(x) \cdot f(y)$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{k}^{\mathrm{x}} \quad(\mathrm{f}(\mathrm{x})>0 \forall \mathrm{x} \in \mathrm{R})$

$\because \mathrm{f}\left(\mathrm{a}_{31}\right)=64 \mathrm{f}\left(\mathrm{a}_{25}\right)$

$\Rightarrow \mathrm{k}^{(\mathrm{a}+30 \mathrm{~d})}=64 . \mathrm{k}^{(\mathrm{a}+24 \mathrm{~d})}$

$\Rightarrow \mathrm{k}^{6 \mathrm{~d}}=64$\ $\Rightarrow \mathrm{k}^{\mathrm{d}}=2$

$\sum_{i=1}^{50} f\left(a_{i}\right)=f\left(a_{1}\right)+f\left(a_{2}\right)+\ldots \ldots+f\left(a_{50}\right)$

$$\begin{aligned} & =k^{a}+k^{a+d}+\ldots .+k^{a+49 d}=\frac{k^{a}\left(k^{50 d}-1\right)}{k^{d}-1} \\ & =k^{a}\left(2^{50}-1\right)=3\left(2^{25}+1\right)(\text { Given }) \\ & \Rightarrow k^{a}=\frac{3}{2^{25}-1} \end{aligned}$$ $$\begin{aligned} & \therefore \sum_{\mathrm{i}=6}^{30} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}}+\mathrm{k}^{\mathrm{a}+6 \mathrm{~d}}+\ldots+\mathrm{k}^{\mathrm{a}+29 \mathrm{~d}} \\ & \quad=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}} \frac{\left(\mathrm{k}^{25 \mathrm{~d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1}=\mathrm{k}^{\mathrm{a}} \cdot\left(\mathrm{k}^{\mathrm{d}}\right)^{5}\left(2^{25}-1\right) \\ & \quad=\frac{3}{2^{25}-1} \cdot 2^{5}\left(2^{25}-1\right)=96 \end{aligned}$$