## Precalculus: Mathematics for Calculus, 7th Edition

$-505$
See p. 855. For the arithmetic sequence $a_{n}=a+(n-1)d$ the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$ is given by either of the following equivalent formulas: 1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad$2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$ ------------------- We see from the given sequence that $a=-10$ $d=-9.9-(-10)=0.1$ The last term is $a_{n}=-0.1$ (we find n): $a_{n}=a+(n-1)d$ $-0.1=-10+0.1(n-1)\qquad/+10$ $9.9=0.1(n-1)\qquad/\times 10$ $99=n-1$ $100=n$ Finally, we find $S_{100}$, (using formula 2) $S_{100}=100(\displaystyle \frac{-10+(-1)}{2})=-505$