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

Solution:

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

This series creates an A.P., with the first term being 6 and the common difference being 6.

a = 6

d = 6

S40 = ?

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

Sn = n/2 [2a +(n – 1)d]

As a result, putting the value of n as n = 40, we get,

S40 = 40/2 [2(6)+(40-1)6]

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

= 20(12+234)

= 20×246

= 4920