Answer
True.
Work Step by Step
The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$ may be found in two ways:
$S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$
$S_{n}=\displaystyle \frac{n}{2}\left(a_{1}+a_{n}\right)$
(theorem)