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

Solution:

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

$\mathrm{S}_{40}=?$

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

$\mathrm{S}_{40}=\frac{40}{2}[2(6)+(40-1) 6]$

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920