Answer
$-505$
Work Step by Step
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. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$
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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)
$S_{100}=100(\displaystyle \frac{-10+(-1)}{2})=-505$
