Answer
1020
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
$a=89$
$d=85-89=-4$
The last term is $a_{n}=13$ (we find n):
$a_{n}=a+(n-1)d$
$13=89-4(n-1)\qquad/-89$
$-76=-4(n-1)\qquad/\div(-4)$
$19=n-1$
$20=n$
And, we find $S_{20}$, (using formula 2)
$S_{20}=20(\displaystyle \frac{89+13}{2})=10(102)=1020$
