Solution;

The two-digit numbers divisible by 3 are 12, 15, 18, …, 99. Clearly, these number are in AP.

Here, a = 12 and d =15 —12 = 3

Let this AP contains n terms. Then,

an = 99

12 + (n – 1) x 3=99         [an=a+(n —1)cl]

3n-F9=99

3n = 99-9 = 90

n = 30

Hence, there are 30 two-digit numbers divisible by 3.