Find the sum of all integers between 100 and 550, which are divisible by 9.

Given:

Integers between 100 and 550, which are divisible by 9.

To do:

We have to find the sum of all integers between 100 and 550, which are divisible by 9.

Solution:

Integers between 100 and 550, which are divisible by 9 are \( 108,117,126, \ldots, 549 \).

The sequence is in A.P.

Here,

\( a=108 \) and \( d=117-108=9 \) \( l=549 \)

We know that,

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

$\Rightarrow 549=108+(n-1) \times 9$

$\Rightarrow 549=108+9n-9$

$\Rightarrow 549-99=9 n$

$\Rightarrow n=\frac{450}{9}=50$

$\therefore n=50$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times 108+(50-1) \times 9]$

$=25[216+49 \times 9]$

$=25(216+441)$

$=25 \times 657$

$=16425$

The sum of all integers between 100 and 550 which are divisible by 9 is $16425$.