The first multiple of 3 between 28 and 112 is 30 , and the last multiple of 3 between 28 and 112 is 111. In order to use Formula 1, the number of terms must be known. \(a_{n}=a_{1}\) \(+(n-1) d\) can be used to find \(n\).
\[
\begin{aligned}
&a_{n}=111, a_{1}=30, d=3 \\
&111=30+(n-1)(3) \\
&81=(n-1)(3) \\
&27=(n-1) \\
&28=n
\end{aligned}
\]
Now, use Formula \(1 .\)
\[
\begin{aligned}
S_{n} &=\frac{n}{2}\left(a_{1}+a_{2}\right) \\
S_{20} &=\frac{28}{2}(30+111) \\
&=14(141) \\
&=1974
\end{aligned}
\]
The sum of the multiples of 3 between 28 and 112 is \(1974 .\)