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

**Solution:**

The sum of the first n terms of an AP is given by S_{n} = n/2 [2a + (n - 1) d] or S_{n} = n/2 [a + l], and the nth term of an AP is a_{n} = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

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

It can be observed that these numbers are making an AP.

Hence,

- First term, a = 6
- Common difference, d = 6
- Number of terms, n = 40

As we know that sum of n terms is given by the formula 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

**Video Solution:**

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

### Class 10 Maths NCERT Solutions - Chapter 5 Exercise 5.3 Question 12:

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

The sum of the first 40 positive integers divisible by 6 is 4920