Question
How many numbers of two digit are divisible by $3?$
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Answer
The first two digit number divisible by $3$ is $12.$
and last two digit number divisible by $3$ is $99.$
So, the required series is $12, 15, 18, ...99.$
Let there be $n$ terms then $n^{th} $ term $= 99$
$\Rightarrow99=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow99=12+(\text{n}-1)3$
$\Rightarrow\text{n}=30$
$30$ two digit number are divisible by $3.$
and last two digit number divisible by $3$ is $99.$
So, the required series is $12, 15, 18, ...99.$
Let there be $n$ terms then $n^{th} $ term $= 99$
$\Rightarrow99=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow99=12+(\text{n}-1)3$
$\Rightarrow\text{n}=30$
$30$ two digit number are divisible by $3.$
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