2016 AMC 10B Problem 18

Suppose the sequence has length $s$ and first term $x$. Then $$345=s\left(x+\frac{s-1}{2}\right),$$ or $$s(2x+s-1)=690.$$ Thus $s$ must be a divisor of $690$, with $x=\frac{\frac{690}{s}-s+1}{2}$ a positive integer. Checking the possible divisor lengths gives $$s=2,3,5,6,10,15,23.$$ These are the only lengths that keep $x$ positive and integral. Therefore there are $7$ representations. Thus, the correct answer is E.