Question

Find the sum of all three digit natural numbers divisible by $3$.

Answer 1

Smallest $3$ digit natural number divisible by $3=102$

Largest $3$ digit natural number divisible by $3=999$

$\therefore 3$ digit number divisible by $3$ are:

$102,105,108,.....999$

this is an $AP$ with $a=102$ and $d=3$

no.of terms

$999=a+(n-1)d$

$999=102+(n-1)3$

$\frac{897}{3}=(n--)\Rightarrow (n-1)=299$

$\therefore n=300$

Sum of all numbers = Sum of $300$ terms of $AP$

$=\frac{300}{2}[2a+)300-1\left)d\right]$

$=\frac{300}{2}[2\times (102)+(299\left)3\right]$

$=\left(150\right)[204+891]$

Sum of all number of series $=\left(150\right)\left[1101\right]=165150$

$\therefore$ Sum of all $3$ digit number divisible by $3$ is $165150$