#### Question

Find the sum of the first 40 positive integers divisible by 3

#### Solution

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

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

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

First 40 positive integers divisible by 3

*n *= number of terms

First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (*a*) = 3

Number of terms (*n*) = 40

Common difference (*d*) = 3

Now, using the formula for the sum of *n* terms, we get

`S_n = 40/2 [2(3) + (40 - 1)3]`

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123)

= 2460

Therefore, the sum of first 40 multiples of 3 is 2460