Question

Find the sum of all natural numbers between 100 and 500 which are divisible by 7.

Answer 1

All the natural numbers less than 100 and divisible by 6 are 6, 12, 18, 24,...,96.

This is an AP in which a = 6, d = (12 - 6) = 6 and l = 96.

Let the number of terms be n.
Then T_n = 96
⇒ a + (n - 1)d = 96
⇒ 6 + (n - 1​) ⨯​ 6 = 96
⇒ 6n = 96
⇒ n = 16

∴ Required sum = $\frac{n}{2}\left(a+l\right)$
= $\frac{16}{2}\left[6+96\right]=8\times 102=816$
Hence, the required sum is 816.