Question

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

Answer 1

The positive integers divisible by 6 are:

6, 12, 18, 24, .......

a = 6, d = 12 – 6 = 6, n = 40

∵ `S_n = n/2 [2a + (n - 1) xx d]`

⇒ `S_40 = 40/2 [2 xx 6 + (40 - 1) xx 6]`

= 20[12 + 39 × 6]

= 20[12 + 234]

= 20(246)

= 4920

Answer 2

The positive integers that are divisible by 6 are

6, 12, 18, 24, ...

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6

S₄₀ = ?

The positive integers that are divisible by 6 are

6, 12, 18, 24, ...

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

`S_40 = 40/2 [2(6) + (40 - 1)6]`

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

= 20(12 + 234)

= 20 × 246

= 4920