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

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

a = 6
d = 6
We have to find \(S_{40}\)

By the formula of sum of n terms, we know,

\(S_n = {{n} \over {2}} [2a +(n – 1)d]\)

Therefore, putting n = 40, we get,

\(S_{40} = {{40} \over {2}} [2(6)+(40-1)6]\)
= \(20[12 + (39)(6)]\)
= \(20(12 + 234)\)
= \(20 × 246\)
= \(4920\)