Question

Find the sum of the integers between 100 and 500 that are divisible by 9.

Answer 1

The first and the last integers in the given range that is divisible by 9 are $108$ and $495$.

$495=55\times 9$

$108=12\times 9$

$\therefore$ Total terms in the given range that are divisible by $9=55-12+1=44$

We know that the sum of $n$ terms of an $A.P.=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}[a+l]$ Where $l$ is the last term, $a$ is the first term. and $d$ is the common difference.

$\therefore Sum=\frac{44}{2}[108+495]=13266$