Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.

Explanation

The numbers between `250` and `1000`
which are divisible by `3` are `252, 255, 258,..........,999`

So,

`T_n​=999=a+(n−1)d`

`=>999=252+(n−1)3`

`=>333=84+n−1`

`=>n=333−83=250`

And,

`S_n=n/2​[a+l]`

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