Question

Find the sum of all three-digit natural numbers which are divisible by $13$.

Answer 1

All three-digit numbers which are divisible by $13$ are $104,117,130,143,...,988.$

This is an AP in which $a=104,d=(117-104)=13$ and $l=988$

Let the number of terms be $n$
Then $T_n=988$
$a+(n-1)d=988$
$104+(n-1)13=988$
$(n-1)13=988-104$
$n-1=\frac{884}{13}$
$n-1=68$
$n=68+1$
$\therefore n=69$
Therefore, the required sum $=\frac{n}{2}(a+l)$
$=\frac{69}{2}(104+988)$
$=\frac{69\times 1092}{2}$
$=69\times 546$
$=37674$
Hence, the required sum is $37674$.