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

First 40 positive integers divisible by 6 are 6, 12, 18, …, 240.

Sum of these numbers forms an arithmetic series 6 + 12 + 18 + … + 240.

Here, first term = a = 6

Common difference = d = 6

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 40 terms of this arithmetic series is given by:

∴ S₄₀ = [2(6) + (40 - 1)(6)]

= 20 [12 + 234]

=20 × 246

= 4920