Q6 of 49 Page 1

Find the number of all three-digit natural numbers which are divisible by 9.

Three-digit numbers = 100, 101, 102, 103, …., 999

Numbers only divisible by 9 = 108, 117, 126, 135, ……, 999


Hence, the series is in A.P.


First term = a = 108


Common difference = 9


Last term = 999


Last term of an A.P. consisting of n terms is given by,


l = a + (n – 1) d


⸫ 999 = 108 + (n – 1) 9


⇒ 9n – 9 = 891


⸫ n = 100


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