Answer
13 terms are in the sum.
Work Step by Step
The nth partial sum od an arithmetic sequence
$S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$
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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$a = 7$ and $d = 3$.
From $S_{n}=325=\displaystyle \frac{n}{2}[2a+(n-1)d]$, we find n:
$325=\displaystyle \frac{n}{2}[14+3(n-1)]$
$325=\displaystyle \frac{n}{2}(11+3n) \quad/\times 2$
$650 =3n^{2}+11n$
$3n^{2}+11n -650=0$
Quadratic formula:
$n=\displaystyle \frac{-11\pm\sqrt{121+4(3)(650}}{2(3)}=\frac{-11\pm 89}{6}$
$n=\displaystyle \frac{78}{6}=13$ or $n=-\displaystyle \frac{50}{3}$
Discarding the negative solution,
13 terms are in the sum.
