12. 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 S40 = ? By the formula of sum of n terms, we know, Sn = n/2 [2a +(n – 1)d] Therefore, putting n = 40, we get, S40 = 40/2 [2(6)+(40-1)6] = 20[12+(39)(6)] = 20(12+234) = 20×246 = 4920