The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be: 
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$
If the angles are in an arithmetic progression, they can be expressed as
$a$, $a+n$, $a+2n$, $a+3n$, and $a+4n$ for some real numbers $a$ and $n$.
Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$.
Adding our expressions for the five angles together, we get $5a+10n=540$.
We now divide by 5 to get $a+2n=108$. It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108\textbf{ (A)}}$.
(In fact, for any arithmetic progression with an odd number of terms,
the middle term is equal to the average of all the terms.)