Question 363369
a1 = 16 and an = 520.  Also, an = a1 + (n-1)d = 16 + (n-1)4 = 4n + 12.
We solve for n.  520 = 4n + 12.
4n = 508, or n = 127.  By the formula for the sum of an arithmetic series,
{{{S = (127/2)*(16 + 520)=34036}}}.