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 \(1974 .\)