[THOUGHT] Let me work through this problem step by step.

First, I need to find the expression (A), the value (B), and the value (C).

From the given conditions:

• $a_1 = 7$
• For $n \geq 2$: $\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10$

For $n \geq 2$: $a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k$

$= \left(\frac{2}{3}a_{n+1} + \frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) + 10\right) - \left(\frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10\right)$

$= \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{6}[(n+1)^2 - n^2] - \frac{1}{6}[(n+1) - n]$

$= \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{6}(2n+1) - \frac{1}{6}$

$= \frac{2}{3}(a_{n+1} - a_n) + \frac{2n+1-1}{6}$

$= \frac{2}{3}(a_{n+1} - a_n) + \frac{2n}{6}$

$= \frac{2}{3}(a_{n+1} - a_n) + \frac{n}{3}$

So (A) = $\frac{n}{3}$

From equation (i): $2a_n + a_{n+1} = 3 \times \frac{n}{3} = n$

Now for (B), substituting $n = 2$: $\sum_{k=1}^{2} a_k = \frac{2}{3}a_2 + \frac{1}{6}(4) - \frac{1}{6}(2) + 10$ $a_1 + a_2 = \frac{2}{3}a_2 + \frac{2}{3} - \frac{1}{3} + 10$ $7 + a_2 = \frac{2}{3}a_2 + \frac{1}{3} + 10$ $7 + a_2 = \frac{2}{3}a_2 + \frac{31}{3}$ $\frac{1}{3}a_2 = \frac{31}{3} - 7 = \frac{31-21}{3} = \frac{10}{3}$ $a_2 = 10$

So (B) = 10

Now for (C): $\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2})$

Using $2a_n + a_{n+1} = n$, we have $2a_{2k+1} + a_{2k+2} = 2k+1$

So $\sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) = 3 + 5 + 7 + 9 + 11 = 35$

Therefore (C) = $a_1 + a_2 + 35 = 7 + 10 + 35 = 52$

Now I need to compute $\frac{p \times q}{f(12)}$:

• $f(n) = \frac{n}{3}$, so $f(12) = \frac{12}{3} = 4$
• $p = 10$ (value B)
• $q = 52$ (value C)

$\frac{p \times q}{f(12)} = \frac{10 \times 52}{4} = \frac{520}{4} = 130$

Let me verify this with Python. [/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve, Rational

Verify the derivation

n = symbols('n')

Expression (A): from a_{n+1} = (2/3)(a_{n+1} - a_n) + (A)

(A) should be the difference in the polynomial parts

(1/6)(n+1)^2 - (1/6)(n+1) + 10 - [(1/6)n^2 - (1/6)n + 10]

= (1/6)[(n+1)^2 - n^2] - (1/6)[(n+1) - n]

= (1/6)(2n+1) - (1/6)(1)

= (2n+1-1)/6 = 2n/6 = n/3

A_expr = n/3 print(f"Expression (A) = n/3") print(f"f(12) = {A_expr.subs(n, 12)}")

Verify (B): a_2

From sum formula with n=2: a_1 + a_2 = (2/3)a_2 + (1/6)(4) - (1/6)(2) + 10

7 + a_2 = (2/3)a_2 + 2/3 - 1/3 + 10

7 + a_2 = (2/3)a_2 + 1/3 + 10

a2 = symbols('a2') eq = Eq(7 + a2, Rational(2,3)*a2 + Rational(1,6)*4 - Rational(1,6)*2 + 10) a2_val = solve(eq, a2)[0] print(f"a_2 = {a2_val}")

Verify the recurrence 2a_n + a_{n+1} = n

From a_{n+1} = (2/3)(a_{n+1} - a_n) + n/3

a_{n+1} - (2/3)a_{n+1} + (2/3)a_n = n/3

(1/3)a_{n+1} + (2/3)a_n = n/3

Multiply by 3: a_{n+1} + 2a_n = n

So 2a_n + a_{n+1} = n

print("Recurrence: 2a_n + a_{n+1} = n")

Compute (C)

sum_{k=1}^{12} a_k + sum_{k=1}^{5} a_{2k+1}

= a_1 + a_2 + sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2})

Using 2a_n + a_{n+1} = n with n = 2k+1:

2a_{2k+1} + a_{2k+2} = 2k+1

sum_part = sum(2*k + 1 for k in range(1, 6)) print(f"sum of (2k+1) for k=1 to 5: {sum_part}")

C_val = 7 + a2_val + sum_part print(f"(C) = a_1 + a_2 + {sum_part} = {C_val}")

Final answer

p = a2_val # (B) q = C_val # (C) f_12 = 12/3 # f(12)

result = (p * q) / f_12 print(f"\np = {p}") print(f"q = {q}") print(f"f(12) = {f_12}") print(f"p * q / f(12) = {result}") [/PYTHON]

[THOUGHT] The Python calculation confirms my work:

• f(n) = n/3, so f(12) = 4
• p (value B) = 10
• q (value C) = 52
• p × q / f(12) = 10 × 52 / 4 = 520 / 4 = 130 [/THOUGHT]

\boxed{130}