[THOUGHT] Let me analyze this problem step by step.

Given:

1. $f: \mathbb{R} \to \mathbb{R}$ with $f(x) > 0$ for all $x$
2. $f(x + y) = f(x)f(y)$ for all $x, y$

This functional equation with positive values means $f(x) = e^{cx}$ for some constant $c$, or equivalently $f(x) = b^x$ for some base $b > 0$.

Let's say $f(x) = b^x$ for some $b > 0$.

Since $a_1, a_2, \ldots, a_{50}$ are in arithmetic progression with common difference $d$: $a_i = a_1 + (i-1)d$

So $f(a_i) = b^{a_i} = b^{a_1 + (i-1)d} = b^{a_1} \cdot (b^d)^{i-1}$

Let $r = b^d$ and $A = b^{a_1} = f(a_1)$. Then $f(a_i) = A \cdot r^{i-1}$.

This means $f(a_1), f(a_2), \ldots, f(a_{50})$ form a geometric progression with first term $A$ and common ratio $r$.

From $f(a_{31}) = 64 f(a_{25})$: $A \cdot r^{30} = 64 \cdot A \cdot r^{24}$ $r^6 = 64 = 2^6$ $r = 2$

From $\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1)$: $A \cdot \frac{r^{50} - 1}{r - 1} = 3(2^{25} + 1)$ $A \cdot \frac{2^{50} - 1}{2 - 1} = 3(2^{25} + 1)$ $A \cdot (2^{50} - 1) = 3(2^{25} + 1)$ $A = \frac{3(2^{25} + 1)}{2^{50} - 1} = \frac{3(2^{25} + 1)}{(2^{25} - 1)(2^{25} + 1)} = \frac{3}{2^{25} - 1}$

Now I need to find $\sum_{i=6}^{30} f(a_i)$: $\sum_{i=6}^{30} f(a_i) = A \cdot \sum_{i=6}^{30} r^{i-1} = A \cdot \sum_{j=5}^{29} 2^j = A \cdot 2^5 \cdot \frac{2^{25} - 1}{2 - 1} = A \cdot 32 \cdot (2^{25} - 1)$

$= \frac{3}{2^{25} - 1} \cdot 32 \cdot (2^{25} - 1) = 3 \cdot 32 = 96$

Let me verify this with Python. [/THOUGHT] [PYTHON]

Let's verify the solution

r = 2 (common ratio)

r = 2

A = 3 / (2^25 - 1)

A = 3 / (2**25 - 1)

Verify sum from i=1 to 50

total_sum = sum(A * r**(i-1) for i in range(1, 51)) print(f"Sum from i=1 to 50: {total_sum}") print(f"Expected: 3*(2^25 + 1) = {3 * (2**25 + 1)}")

Verify f(a_31) = 64 * f(a_25)

f_a31 = A * r30 f_a25 = A * r24 print(f"f(a_31) / f(a_25) = {f_a31 / f_a25}") print(f"Expected: 64")

Calculate sum from i=6 to 30

sum_6_to_30 = sum(A * r**(i-1) for i in range(6, 31)) print(f"Sum from i=6 to 30: {sum_6_to_30}")

Let's also compute it symbolically

from sympy import *

A * sum of r^(i-1) for i from 6 to 30

= A * (r^5 + r^6 + ... + r^29)

= A * r^5 * (1 + r + ... + r^24)

= A * r^5 * (r^25 - 1) / (r - 1)

= A * 32 * (2^25 - 1) / 1

= (3 / (2^25 - 1)) * 32 * (2^25 - 1)

= 3 * 32 = 96

symbolic_result = Rational(3, 225 - 1) * 32 * (225 - 1) print(f"Symbolic result: {symbolic_result}") [/PYTHON]

[THOUGHT] The Python verification confirms all our calculations:

1. The sum from i=1 to 50 equals 3(2^25 + 1) = 100663299 ✓
2. f(a_31)/f(a_25) = 64 ✓
3. The sum from i=6 to 30 equals 96 ✓

The answer is 96. [/THOUGHT]

\boxed{96}