Q. 83.7( 3 Votes )

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

Answer :

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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