Question

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

Answer 1

The positive integers that are divisible by 6 are;
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
$S_{40}=?S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{40}=\frac{40}{2}\left[2\right(6)+(40-1\left)6\right]=20[12+(39\left)\right(6\left)\right]=20(12+234)=20\times 246=4920$