Question

Find the sum of all natural numbers from $100$ to $300$.
Which are divisible by $5$.
Which are divisible by $6$.
Which are divisible by $5$ and $6$.

Answer 1

Numbers divisible by $5$ between $100$ and $300$ are

$105,110,----295$

$a=100,d=5$ last term $l=295$

$t_n=a+(n-1)d\Rightarrow 295=100+(n-1)\times 5$
$\Rightarrow \frac{195}{5}=n-1$
$\Rightarrow 39=n-1$
$\Rightarrow n=39+1=40$
$\therefore Sum=\frac{n}{2}(a+l)=\frac{40}{2}(105+295)$
$=20\times 400=8000$

Numbers divisible by $6$ are

$102,108,\mathrm{114...294}$

$a=102,l=294$

and $294=102+(n+1)\times 6\Rightarrow 294-102=(n-1)\times 6$
$\Rightarrow (n-1)\times 6=192$
$\Rightarrow n-1=32$
$\Rightarrow n=32+1=33$
$\therefore sum=\frac{33}{2}(102+294)=\frac{33}{2}\times 396=6534$

Numbers divisible by both $5$ and $6$ are

$120,150,180,210,240,270a=120,d=30$
$l=270$

Also $n=6$

$\therefore sum=\frac{6}{2}(120+270)=3\times 390=1170$