Question

Find the number of all three-digit natural numbers which are divisible by $9$.

Answer 1

The three digit numbers which are divisible by 9 are

$108,117,126,...\dots \dots ..999$

which forms an A.P

the first term of this A.P is $a_1=108$

second term of this A.P is $a_2=117$

common difference of this A.P is

$d=a_2-a_1=117-108=9$

the nth term of this A.P is given by

$a_n=a_1=(n-1)d$

$⟹a_n=108+(n-1)9$

$⟹a_n=108+9n-9$

$⟹a_n=99+9n......eq\left(1\right)$

since last term of this A.P is $a_n=999$

hence for finding the number of terms (n) in this A.P put $a_n=999$ in eq(1)

$⟹999=99+9n$

$⟹9n=999-99$

$⟹9n=900$

$⟹n=\frac{900}{9}$

$⟹n=100$

hence there are 100 three digit terms which are divisible by $9$.