Find the sum of the integers between 100 and 200 that are divisible by 9

To do:

We have to find the sum of all integers between 100 and 200 that are

(i) divisible by 9

(ii) not divisible by 9

Solution:

(i) Integers between 100 and 200, that are divisible by 9 are \( 108,117,126, \ldots, 198 \).

The sequence is in A.P.

Here,

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

\( l=198 \)

We know that,

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

$\Rightarrow 198=108+(n-1) \times 9$

$\Rightarrow 198=108+9n-9$

$\Rightarrow 198-99=9 n$

$\Rightarrow n=\frac{99}{9}=11$

Therefore,

$n=11$

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

$S_{11}=\frac{11}{2}[2 \times 108+(11-1) \times 9]$

$=\frac{11}{2}[216+10 \times 9]$

$=11(108+45)$

$=11 \times 153$

$=1683$

The sum of all integers between 100 and 200 that are divisible by 9 is $1683$.    

(ii) Sum of the integers between 100 and 200, which are not divisible by 9 $=$ Sum of the integers between 100 and 200 $-$ Sum of the integers between 100 and 200, which are divisible by 9

Integers between 100 and 200 are \( 101,102,103, \ldots, 199 \).

The sequence is in A.P.

Here,

\( a=101 \) and \( d=102-101=1 \) \( l=199 \)

We know that,

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

$\Rightarrow 199=101+(n-1) \times 1$

$\Rightarrow 199=101+n-1$

$\Rightarrow 199-100=n$

$\Rightarrow n=99$

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

$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$

$=\frac{99}{2}[202+98 \times 1]$

$=\frac{99}{2}(202+98)$

$=99 \times 150$

$=14850$

From (i)

The sum of all integers between 100 and 200 that are divisible by 9 is $1683$.    

Therefore,

Sum of all integers between 100 and 200, which are not divisible by 9 $=14850-1683$

$=13167$

The sum of all integers between 100 and 200 which are not divisible by 9 is $13167$.