Question 12

The functional equation $f(x+y) = f(x)f(y)$ with $f(x) > 0$ implies $f(x) = k^x$ for some $k > 0$. Since $a_i$ is an AP, $a_i = a_1 + (i-1)d$. Then $f(a_i) = k^{a_1 + (i-1)d} = f(a_1) \cdot (k^d)^{i-1}$. Let $f(a_1) = A$ and $k^d = r$. Then $f(a_i) = A r^{i-1}$, which is a GP. Given $f(a_{31}) = 64 f(a_{25}) \implies A r^{30} = 64 A r^{24} \implies r^6 = 64 \implies r = 2$ (since $f(x)>0$). Now, $\sum_{i=1}^{50} f(a_i) = A \frac{r^{50}-1}{r-1} = A(2^{50}-1) = 3(2^{25}+1)$. $A(2^{25}-1)(2^{25}+1) = 3(2^{25}+1) \implies A = \frac{3}{2^{25}-1}$. We need to find $\sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} A r^{i-1} = A(r^5 + r^6 + \dots + r^{29})$. This is a GP with $25$ terms: $A \cdot r^5 \frac{r^{25}-1}{r-1} = A \cdot 2^5 (2^{25}-1)$. Substitute $A$: $\frac{3}{2^{25}-1} \cdot 32 \cdot (2^{25}-1) = 3 \times 32 = 96$.