The arithmetic mean (average) of the first $n$ positive integers is:
$\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2}$
The sum of the first $n$ positive integers is the same as the nth triangular number, which can be expressed as $\frac{(n)(n+1)}{2}$. Since the question is asking for the arithmetic mean, the whole sum is divided by $n$, thus giving us $\frac {\frac{(n)(n+1)}{2}}{n}$, which then simplifies to $\textbf{(E)} \frac{n+1}{2}$