Let $\{a_n\}$ be a sequence satisfying the following conditions.\begin{center}\fbox{\begin{minipage}{0.9\textwidth}

• $a_1 = 7$,
• For every natural number $n \geq 2$, $\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10.$ \end{minipage}}\end{center}Consider the following work to compute$\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1}.$\begin{center}\fbox{\begin{minipage}{0.9\textwidth}For every natural number $n \geq 2$,$a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k,$so$a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})},$and, after rearranging this expression,$2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i)}.$From$\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \quad (n \geq 2),$substituting $n = 2$ into both sides gives$a_2 = \boxed{(\text{B})} \qquad \cdots\cdots \text{(ii)}.$By (i) and (ii),\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})$$$$= \boxed{(\text{C})}.\end{minipage}}\end{center}Let $f(n)$ be the expression corresponding to \boxed{(\text{A})}, and let $p$ and $q$ be the numbers corresponding to \boxed{(\text{B})} and \boxed{(\text{C})}, respectively. Compute$\dfrac{p \times q}{f(12)}.$[4 points]\