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

Advertisement Remove all ads

#### Solution

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

S_{40} =?

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

S_{40} =?

`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

Concept: Sum of First n Terms of an AP

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads