## Precalculus (6th Edition) Blitzer

a. $a_n=-5(-2)^{n-1}$ b. $640$ c. $1705$
a. Based on the given sequence, we have $a_1=-5, r=\frac{10}{-5}=-2$ hus $a_n=a_1(r)^{n-1}=-5(-2)^{n-1}$ b. For $n=8$, we have $a_8=-5(-2)^{8-1}=640$ c. For $n=10$, the sum is $S_n=\frac{a_1(1-r^n)}{1-r}$ Thus $S_{10}=\frac{-5(1-(-2)^{10})}{1-(-2)}=1705$