Question

Find the sum of first $15$ multiples of $8$.

Answer 1

The first $15$ multiples of $8$ are $8,16,24,32,...$

This is an AP in which,

First term $\left(a\right)=8$

Common difference $\left(d\right)=t_2-t_1=16-8=8$

For first $15$ terms, $\left(n\right)=15$

We know that, The sum of first $n$ terms in A.P is,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow S_{15}=\frac{15}{2}\left(2\right(8)+(15-1\left)8\right)$

$\Rightarrow S_{15}=\frac{15}{2}(16+112)$

$\Rightarrow S_{15}=\frac{15}{2}\left(128\right)$

$\Rightarrow S_{15}=15\times 64$

$\Rightarrow S_{15}=960$

Hence,sum of first $15$ multiples of $8$ is $960$.