Answer
$13$ terms
Work Step by Step
Let's note by $\{a_n\}$ the given arithmetic sequence. Its elements are:
$$\begin{cases}
a_1=7\\
d=3.
\end{cases}$$
The $n$th term is:
$$a_n=a_1+(n-1)d=7+(n-1)(3)=7+3n-3=3n+4.$$
In order to find the number of terms so that $S_n=325$, we use the formula:
$$S_n=\dfrac{n(a_1+a_n)}{2}$$
which we solve for $n$:
$$\begin{align*}
325&=\dfrac{n(7+3n+4)}{2}\\
650&=n(3n+11)\\
3n^2+11n-650&=0\\
(n-13)(3n+50)&=0\\
n-13=0&\text{ or }3n+50=0\\
n=13&\text{ or } n=-\dfrac{50}{3}.
\end{align*}$$
As $n$ must be a positive integer, the only solution is
$$n=13.$$
So $13$ terms are needed so that the sum is $325$.
