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

Solution:

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

This is an AP with first term 6 and the common difference 6.

a = 6

d = 6

S₄₀ = ?

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

S_n = 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