Question

Find the sum of all 3- digit natural numbers, which are multiples of $11$.

Answer 1

First three digit number divisible by $11$ is $110$

Last three digit numbers divisible by $11$ is $990$

$\therefore a=110$ $a_n=990$ $d=11$ $x=2$

$\therefore$ No. of $=2$

Terms

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

$990=110+(n-1).11$

$\frac{880}{11}=x-1$

$80=x-1$

$x=81$

$\therefore$ sum of all terms of A.P is given by

$S_n=[a+a_n]\times \frac{n}{2}$

$=[110+990]\times \frac{81}{2}$

$=1100\times \frac{81}{2}$

$=550\times 81$

$=44550$