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Find the sum of the multiples of 3 between 28 and $112 .$
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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 $1974 .$
by Diamond (78,219 points)

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