Question

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

Answer 1

Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999.

This is an A.P. with first term a = 252, common difference = 3 and last term = 999.

Let there be n terms in this A.P. Then,

⇒ a_n = 999

⇒ a + (n – 1)d = 999

⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250

∴ Required sum = `S_n = \frac { n }{ 2 } [a + l]`

`= \frac { 250 }{ 2 } [252 + 999] = 156375`