Question

Find the sum of the first $40$ positive integers divisible by $6$.

Answer 1

The first $40$ positive integers that are divisible by $6$ are $6,12,18,24\dots$

$a=6$ and $d=6$.

We need to find $S_{40}$

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

$S_{40}=\frac{40}{2}\left[2\right(6)+(40-1\left)6\right]$

$=20[12+(39\left)6\right]$

$=20(12+234)$

$=20\times 246$

$=4920$