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

The first 40 positive integers divisible by 6 are 6, 12, 18, …, 240.
Here, a = 6, d = 12 – 6 = 6 and n = 40.
The sum of n terms of an AP is Given by
S_n = n/2[2a + (n -1)d]
⇒ S_40 = 10/2[2(6) + (40 -1)(6)]
= 20[12 + 234]
= 20(246) = 4920
Hence, the sum of the first 40 positive integers divisible by 6 is 4920.