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

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

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

a = 6
d = 6
We have to find $$S_{40}$$

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

$$S_n = {{n} \over {2}} [2a +(n – 1)d]$$

Therefore, putting n = 40, we get,

$$S_{40} = {{40} \over {2}} [2(6)+(40-1)6]$$
= $$20[12 + (39)(6)]$$
= $$20(12 + 234)$$
= $$20 × 246$$
= $$4920$$