Let \(\mathbb{R}\) denote the set of all real numbers. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function such that \(f(x)>0\) for all \(x \in \mathbb{R}\), and \(f(x+y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\).
Let the real numbers \(\mathrm{a}_1, \mathrm{a}_2 \ldots, \mathrm{a}_{50}\) be in an arithmetic progression. If \(f\left(\mathrm{a}_{31}\right)=64 f\left(\mathrm{a}_{25}\right)\), and \(\sum_{i=1}^{50} f\left(a_i\right)=3\left(2^{25}+1\right)\) then the value of \(\sum_{i=6}^{30} f\left(a_i\right)\) is ________

\(\begin{aligned} & \because f(x+y)=f(x) \cdot f(y) \\ & \Rightarrow f(x)=k^x \quad(f(x)>0 \forall x \in R) \\ & \because f\left(a_{31}\right)=64 f\left(a_{25}\right)\end{aligned}\)
\(\begin{aligned} & \Rightarrow \mathrm{k}^{(\mathrm{a}+30 \mathrm{~d})}=64 \cdot \mathrm{k}^{(\mathrm{a}+24 \mathrm{~d})} \\ & \Rightarrow \mathrm{k}^{6 \mathrm{~d}}=64 \\ & \Rightarrow \mathrm{k}^{\mathrm{d}}=2\end{aligned}\)
\(\sum_{\mathrm{i}=1}^{50} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{f}\left(\mathrm{a}_1\right)+\mathrm{f}\left(\mathrm{a}_2\right)+\ldots \ldots+\mathrm{f}\left(\mathrm{a}_{50}\right)\)
\(\begin{aligned} & =\mathrm{k}^{\mathrm{a}}+\mathrm{k}^{\mathrm{a}+\mathrm{d}}+\ldots+\mathrm{k}^{\mathrm{a}+49 \mathrm{~d}}=\frac{\mathrm{k}^{\mathrm{a}}\left(\mathrm{k}^{50 \mathrm{~d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1} \\ & =\mathrm{k}^{\mathrm{a}}\left(2^{50}-1\right)=3\left(2^{25}+1\right)(\text { Given }) \\ & \Rightarrow \mathrm{k}^{\mathrm{a}}=\frac{3}{2^{25}-1}\end{aligned}\)
\(\begin{aligned} & \therefore \sum_{\mathrm{i}=6}^{30} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}}+\mathrm{k}^{\mathrm{a}+6 \mathrm{~d}}+\ldots+\mathrm{k}^{\mathrm{a}+29 \mathrm{~d}} \\ & \quad=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}} \frac{\left(\mathrm{k}^{25 \mathrm{~d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1}=\mathrm{k}^{\mathrm{a}} \cdot\left(\mathrm{k}^{\mathrm{d}}\right)^5\left(2^{25}-1\right) \\ & \quad=\frac{3}{2^{25}-1} \cdot 2^5\left(2^{25}-1\right)=96\end{aligned}\)