Question

Sum of numbers from $1$ to $140$, which are divisible by $4$= ?

Answer 1

The natural numbers from 1 to 140 that are divisible by 4 are as follows : $4,8,12,16,..............,140$

These numbers from an A.P. with $a=4,d=4$

Let, 140 be the nth term of A.P.

$\therefore T_n=140$

$T_n=a+(n–1)d$

$140=4+(n–1)4$

$140=4+4n–4$

$140=4n$

$n=35$

$S_n=\frac{n}{2}[2a+(n–1\left)d\right]$

$S_{35}=\frac{35}{2}\left[2\right(4)+(35–1\left)4\right]$

$S_{35}=\frac{35}{2}[8+34(4\left)\right]$

$S_{35}=\frac{35}{2}[8+136]$

$S_{35}=2520$