Question

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

• A. 156375
• B. 156324
• C. 147687
• D. 156342

Answer 1

The correct option is A

156375

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) $\times$ 3 = 999

n = 250

Required sum = $S_n$ = ($\frac{n}{2}$) [a + l]

= ($\frac{250}{2}$) [252 + 999] = 156375