Very-Short and Short-Answer Questions

The three-digit natural numbers divisible by 9 are 108, 117, 126, ..., 999.

Clearly, these number are in AP.

Here, a = 108 and d = 117 − 108 = 9

$a_{n}=999$

$\Rightarrow 108+(n-1) \times 9=999 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 9 n+99=999$

$\Rightarrow 9 n=999-99=900$

$\Rightarrow n=100$

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

Let this AP contains n terms. Then,