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

The positive integers divisible by 6 are 6, 12, 18, ... .

This is an AP with a = 6 and d = 6.

Also, n = 40           (Given)

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get

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

$=20(12+234)$

$=20 \times 246$

$=4920$

Hence, the required sum is 4920.