The sequence $\left\{ a _ { n } \right\}$ satisfies the following conditions.
\begin{itemize}
  \item $a _ { 1 } = 7$
  \item For natural numbers $n \geq 2$,
\end{itemize}
$$\sum _ { k = 1 } ^ { n } a _ { k } = \frac { 2 } { 3 } a _ { n } + \frac { 1 } { 6 } n ^ { 2 } - \frac { 1 } { 6 } n + 10$$

The following is the process of finding the value of $\sum _ { k = 1 } ^ { 12 } a _ { k } + \sum _ { k = 1 } ^ { 5 } a _ { 2 k + 1 }$.

For natural numbers $n \geq 2$, since $a _ { n + 1 } = \sum _ { k = 1 } ^ { n + 1 } a _ { k } - \sum _ { k = 1 } ^ { n } a _ { k }$,
$$a _ { n + 1 } = \frac { 2 } { 3 } \left( a _ { n + 1 } - a _ { n } \right) + \text { (가) }$$
and simplifying this equation gives
$$2 a _ { n } + a _ { n + 1 } = 3 \times \text { (가) } \quad \cdots \cdots \text { (ㄱ) }$$
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 $n = 2$ gives
$$a _ { 2 } = \text { (나) }$$
By (ㄱ) and (ㄴ),
$$\begin{aligned}
\sum _ { k = 1 } ^ { 12 } a _ { k } + \sum _ { k = 1 } ^ { 5 } a _ { 2 k + 1 } & = a _ { 1 } + a _ { 2 } + \sum _ { k = 1 } ^ { 5 } \left( 2 a _ { 2 k + 1 } + a _ { 2 k + 2 } \right) \\
& = \text { (다) }
\end{aligned}$$
Let $f ( n )$ be the expression that fits in (가), and let $p$ and $q$ be the numbers that fit in (나) and (다), respectively. Find the value of $\frac { p \times q } { f ( 12 ) }$. [4 points]