Answer
$310.5$
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
$a=-3,$
$d=\displaystyle \frac{3}{2}-0=\frac{3}{2}$
The last term is $a_{n}=30$ (we find n):
$a_{n}=a+(n-1)d$
$30=-3+\displaystyle \frac{3}{2}(n-1)$
$33=\displaystyle \frac{3}{2}(n-1)\qquad/\times\frac{2}{3}$
$22=n-1$
$23=n$
And, we find $S_{23}$, (using formula 2)
$S_{23}=23(\displaystyle \frac{-3+30}{2})=\frac{621}{2}=310.5$
