Question

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

Answer 1

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

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

Common difference $d=3$

Let the total number of 3 digit number be $n$

Therefore,

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

substituting the values

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

$999-102=(n-1)\times 3$

$897=(n-1)\times 3$

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

$n-1=299$

$n=300$

Now,

Sum of numbers in sequence$=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{300}{2}[2\times 102+(300-1)\times 3]$

$=165150$

Therefore sum of all 3 digit number which are divisible by $3$ is $165150$.