P2859
By hypothesis, $a$ satisfies
\begin{equation}
N a = \sum_{n = 1}^N a = \sum_{n = 1}^N x_n
\end{equation}
Diving each side by the positive integer $N \geq 1$, we obtain
\begin{equation}
a
= \frac{1}{N} \sum_{n = 1}^N x_n
\end{equation}
$\square$
