Solution:

The three-digit numbers divisible by 9 are 108, 117, 126,…., 999. Clearly, these number are in AP.

Here. a= 108 and d= 117 — 108 = 9

Let this AP contains n terms. Then.

a_n = 999

=.108+(n – 1) x 9 =999                    [a_n = a + (n -1)d]

9n +99 =999

9n = 999 —99 = 900

n=100

Hence: there are 100 three-digit numbers divisible by 9.