## Precalculus (6th Edition) Blitzer

$6820$
we have: $a_1=-20$ and $r=-2$ For a geometric sequence, we have $S_n=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{10}=\dfrac{-20(1-(-2)^{10})}{1-2}$ or, $=\dfrac{20460}{3}$ Thus, $S_{10}=6820$