Question

Find the sum of all $3$ digit natural numbers, which are divisible by $9$

Answer 1

$3$ digit number starts from $100$ and ends with $999$. From this sequence we have to find number of terms which are divisible by $9$ and those terms are as follows:

$108,117,126,.....,999$

In the above sequence, the first term is $a_1=108$, second term is $a_2=117$ and the $n$th term is $T_n=999$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

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

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, with $a=108,d=9$ and $T_n=999$, we have

$T_n=a+(n-1)d\Rightarrow 999=108+(n-1)9\Rightarrow 999=108+9n-9\Rightarrow 999=99+9n\Rightarrow 9n=999-99\Rightarrow 9n=900\Rightarrow n=\frac{900}{9}\Rightarrow n=100$

We also know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of series, substitute $n=100,a=108$ and $d=9$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{100}=\frac{100}{2}\left[\right(2\times 108)+(100-1\left)9\right]=50[216+(99\times 9\left)\right]=50(216+891)=50\times 1107=55350$

Hence, the sum is $55350$.