Find the sum of all three–digit numbers divisible by 3.

Formula used.

S_n = [a + l]

a_n = a + (n–1)d

⇒ The 1^st 3 digit number divisible by 3 is 102

For the last 3 digit number divisible by 3

Divide 999 by 3

It gets completely divisible by 3

⇒ The last 3 digit number divisible by 3 is 999

a_n = a + (n–1)d

999 = 102 + (n–1)3

999 = 102–3 + 3n

3n = 999 – 99

n = = 300

S_n = [a + l]

= [102 + 999]

= 150×1101

= 165150