Question

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

Answer 1

The multiples of $8$ are $8,16,24,32\dots$
These are in an A.P., having first term as $8$ and common difference as $8$.
Therefore, $a=8,d=8$, $S_{15}=?$

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

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

$=\frac{15}{2}[16+(14\left)\right(8\left)\right]$

$=\frac{15\times 128}{2}$

$=960$