Question

Find the sum of all three digit natural numbers which are divisible by 7.

Answer 1

First no is $105$ and last no is $994$

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

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

$994=105+(n-1)7$

$994=105+7n-7$

$994+7-105=7n$
$896=7n$

$\therefore n=128$

Now, as we know

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

$\therefore$ $S_{128}=\frac{128}{2}[210+889]$......putting $(n=128)$

$=64\times 1099=70336$