Question

Find the sum of all three digit numbers between $100$ and $300$ divisible by $6$.

Answer 1

Numbers between 100 and 300 which are divisible by 6 are form 102,108.....294.

There are in from of AP with first term 102, last term = 294

a=102 d=6 l=294

$294=102+(n-1)6$

$(n-1)6=192\Rightarrow n-1=32$

n=33

$\therefore$ sum of n terms of AP

$=\frac{n}{2}[a+l]$ a-first term

$=\frac{33}{2}[102+294]=6534$