[THOUGHT] We are given a functional equation: f(x+y) = f(x)f(y) for all real x, y, and f(x) > 0 for all x. This is a classic Cauchy exponential equation. Since f is positive, we can write f(x) = e^{g(x)} for some function g. Then f(x+y) = f(x)f(y) implies e^{g(x+y)} = e^{g(x)} e^{g(y)} = e^{g(x)+g(y)}. Taking logs, we get g(x+y) = g(x) + g(y). If we assume continuity or measurability, then g is linear: g(x) = cx for some constant c. Thus f(x) = e^{cx} = b^x where b = e^c > 0. But we are not given continuity; however, the condition f(x) > 0 and the functional equation implies f is an exponential function: f(x) = b^x for some b > 0. Actually, we can show: Let f(1) = b > 0. Then for integer n, f(n) = f(1)^n = b^n. For rationals, similarly. Since we are dealing with real numbers in arithmetic progression, we can assume f(x) = b^x for some b > 0. But we must verify that the given conditions are consistent with that form.

Given a1, a2, ..., a50 are in arithmetic progression. Let the common difference be d. So a_i = a_1 + (i-1)d.

We have f(a_{31}) = 64 f(a_{25}). Using f(x) = b^x, this becomes b^{a_{31}} = 64 b^{a_{25}} => b^{a_{31} - a_{25}} = 64. But a_{31} - a_{25} = 6d. So b^{6d} = 64 = 2^6. Thus (b^d)^6 = 2^6 => b^d = 2 (since b>0). So b^d = 2.

Now we have sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1). f(a_i) = b^{a_i} = b^{a_1 + (i-1)d} = b^{a_1} * (b^d)^{i-1} = b^{a_1} * 2^{i-1}. Let A = b^{a_1}. Then f(a_i) = A * 2^{i-1}.

Thus sum_{i=1}^{50} f(a_i) = A * sum_{i=1}^{50} 2^{i-1} = A * (2^{50} - 1)/(2 - 1) = A * (2^{50} - 1).

Given this equals 3(2^{25} + 1). So A * (2^{50} - 1) = 3(2^{25} + 1). Note 2^{50} - 1 = (2^{25} - 1)(2^{25} + 1). So A * (2^{25} - 1)(2^{25} + 1) = 3(2^{25} + 1). Assuming 2^{25}+1 ≠ 0, we can cancel: A * (2^{25} - 1) = 3. Thus A = 3/(2^{25} - 1).

Now we need sum_{i=6}^{30} f(a_i). f(a_i) = A * 2^{i-1}. So sum = A * sum_{i=6}^{30} 2^{i-1} = A * sum_{j=5}^{29} 2^j (where j = i-1). Compute sum_{j=5}^{29} 2^j = 2^5 + 2^6 + ... + 2^29 = 2^5 * (2^{25} - 1)/(2 - 1) = 32 * (2^{25} - 1). Wait: number of terms from 5 to 29 inclusive is 25 terms. Indeed sum_{j=5}^{29} 2^j = 2^5 * (2^{25} - 1)/(2-1) = 32*(2^{25} - 1). So sum = A * 32 * (2^{25} - 1). But A = 3/(2^{25} - 1). So product = 3/(2^{25} - 1) * 32 * (2^{25} - 1) = 3 * 32 = 96.

Thus answer is 96.

We should verify with Python to ensure no arithmetic mistakes. We'll compute using symbolic or numeric values.

[/THOUGHT] [PYTHON] import sympy as sp

Define variables

b, d, a1 = sp.symbols('b d a1', positive=True)

We derived b^d = 2

Let's compute A = b^{a1}

A = b**a1

From condition sum f(a_i) = 3(2^{25}+1)

f(a_i) = b^{a1 + (i-1)d} = b^{a1} * (b^d)^{i-1} = A * 2^{i-1}

Sum_{i=1}^{50} A * 2^{i-1} = A * (2^{50} - 1)

Equation: A * (250 - 1) = 3*(225 + 1)

Solve for A

A_val = 3*(225 + 1) / (250 - 1) print("A =", A_val.simplify())

Should be 3/(2**25 - 1)

print("Simplified A =", (3/(2**25 - 1)).simplify())

Now compute sum_{i=6}^{30} f(a_i) = A * sum_{i=6}^{30} 2^{i-1}

sum_powers = sum(2**(i-1) for i in range(6, 31)) # i from 6 to 30 inclusive print("sum_{i=6}^{30} 2^{i-1} =", sum_powers)

Compute expected: 32*(2**25 - 1)

expected_sum = 32 * (2**25 - 1) print("Expected sum =", expected_sum) print("Are they equal?", sum_powers == expected_sum)

Compute total sum = A * sum_powers

total = A_val * sum_powers print("Sum_{i=6}^{30} f(a_i) =", total.simplify())

Should be 96

print("As integer:", int(total)) [/PYTHON]

[THOUGHT] The Python code had an error because I tried to call .simplify() on a float. However, the calculation before the error shows that A_val = 3*(225 + 1) / (250 - 1) is a numeric value. The sum of powers from i=6 to 30 of 2^{i-1} is computed correctly. The product A_val * sum_powers should yield 96. Since the code didn't complete due to the attribute error, I can recompute manually: A = 3/(2^25 - 1), sum_powers = 32*(2^25 - 1), product = 332 = 96. This matches my earlier reasoning. The Python output before the error shows sum_powers = 1073741792? Actually, let's check: 2^30 is 1073741824, so the sum from 2^5 to 2^29 is indeed 32(2^25 - 1) = 1073741792. Multiplying by A = 3/(2^25 - 1) gives 3*32 = 96. So the answer is 96.

[/THOUGHT] \boxed{96}