# Find the sum of all 3 digit natural numbers, which are divisible by 9.

Series of three digit numbers divisible by 9 is:

108, 117,………………………………999

In the A.P.

First term = 108

Last term = 999

Common difference = 9

Nth term = a + (n–1) d

999 = 108 + (n–1)9

891 = (n–1)9

99 = (n–1)

n = 100

Sum of terms =

Sum of terms =

Sum of terms =

Sum of terms = 55350

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