Sum of Terms in an Exponential Functional Sequence

Given the functional equation $f(x+y) = f(x)f(y)$ with $f(x) > 0$ for all $x \in \mathbb{R}$, the sequence $f(a_i)$ forms a geometric progression. Let the common difference of the arithmetic progression $a_1, a_2, \dots, a_{50}$ be $d$.

Let $A = f(a_1)$ and the common ratio be $r = f(d) > 0$. Then, for any $i \ge 1$:

$$f(a_i) = A r^{i-1}$$

Using the given condition $f(a_{31}) = 64 f(a_{25})$:

$$A r^{30} = 64 A r^{24} \implies r^6 = 64 \implies r = 2$$

From the given sum of the first $50$ terms:

$$\sum_{i=1}^{50} f(a_i) = A \sum_{i=1}^{50} 2^{i-1} = A (2^{50} - 1) = A (2^{25} - 1)(2^{25} + 1)$$

Given that this sum equals $3(2^{25} + 1)$, we obtain:

$$A(2^{25} - 1)(2^{25} + 1) = 3(2^{25} + 1) \implies A(2^{25} - 1) = 3$$

Now, we evaluate the required sum:

$$\sum_{i=6}^{30} f(a_i) = \sum_{i=6}^{30} A \cdot 2^{i-1} = A \cdot 2^5 \sum_{k=0}^{24} 2^k = A \cdot 32 (2^{25} - 1)$$

Substituting $A(2^{25} - 1) = 3$:

$$\sum_{i=6}^{30} f(a_i) = 32 \times 3 = 96$$