How many two digit numbers are divisible by $3?$

Given: Two digit numbers.

To do: To find the numbers of  two digit number which are divisible by $3$.

Solution: 

We know, first two digit number divisible by $3$ is $12$ and last two digit number divisible by $3$ is $99$. Thus, we get

$12,\ 15,\ 18,\ ...,\ 99$ which is an AP

Here, $a=12,\ d=3$

Let there be n terms. Then,

$\Rightarrow a_{n}=99$

$\Rightarrow a+(n−1)d=99$

$\Rightarrow 12+(n−1)3=99$

$\Rightarrow n=29+1=30$

Therefore, two digit numbers divisible by $3$ are $30$.