Ans: C
Let $a$ and $d$ be the first term and the common difference of the arithmetic sequence respectively. Then we have
Let $a$ and $d$ be the first term and the common difference of the arithmetic sequence respectively. Then we have
$\left\{ \begin{array}{ll}
a+2d=42 & \ldots \unicode{x2460} \\
a+11d=6 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
9d & = & -36 \\
d & = & -4
\end{array}$
Sub. $d=-4$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
a + 2(-4) & = & 42 \\
a & = & 50
\end{array}$
Therefore, the sum of the first $n$ terms of the sequence
$\begin{array}{cl}
= & \dfrac{n}{2}[ 2(50) +(n-1)(-4)] \\
= & \dfrac{n}{2}(104-4n) \\
= & 52n-4n^2
\end{array}$
