Question

Find out the sum of all natural numbers between $1$ and $145$ which are divisible by $4$.

Answer 1

All natural numbers bwetween $1$ and $145$ which are divisible by $4$ are,

$4,8,12,16,...........,144$ forms an $A.P$

with $a=4$ and $d=4$

Let, $t_n=144$

$⟹a+(n-1)d=144$

$⟹4+(n-1)4=144$

$⟹4(n-1)=140$

$⟹n-1=35$

$\therefore n=36$

Now, $S_n=\frac{n}{2}[a+l]$

$=\frac{36}{2}[4+144]$

$=18\times 148$

$=2664$

Hence, The required sum $=2,664$