Sum of an Arithmetic Progression

The given series is an A.P. with first term $a = -64$, common difference $d = -2$, and last term $a_n = -100$.

Using $a_n = a + (n - 1)d$: $-100 = -64 + (n - 1)(-2) \implies -36 = (n - 1)(-2) \implies n - 1 = 18 \implies n = 19$

The sum $S_n = \frac{n}{2}(a + a_n)$: $S_{19} = \frac{19}{2}(-64 - 100) = \frac{19}{2}(-164) = 19 \times (-82) = -1558$