# Find the sum of the integers

Question:

Find the sum of the integers between 100 and that are

(i) divisible by 9.

(ii) not divisible by 9.

Solution:

(i) The numbers (integers) between 100 and 200 which is divisible by 9 are 108, 117, 126, … 198.

Let n be the number of terms between 100 and 200 which is divisible by 9.

Here, $a=108, d=117-108=9$ and $a_{n}=l=198$

$\because \quad a_{n}=l=a+(n-1) d$

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

$\Rightarrow \quad 90=(n-1) 9$

$\rightarrow$ - $n-1=10$

$\Rightarrow \quad n=11$

$\therefore$ Sum of terms between 100 and 200 which is divisible by 9 ,

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

$\Rightarrow$  $S_{11}=\frac{11}{2}[2(108)+(11-1) 9]=\frac{11}{2}[216+90]$

$=\frac{11}{2} \times 306=11 \times 153=1683$

Hence, required sum of the integers between 100 and 200 that are divisible by 9 is 1683.

(ii) The sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total

numbers between 100 and 200 which is divisible by 9).

Here, $\quad a=101, d=102-101=1$ and $a_{n}=l=199$

$\because \quad a_{n}=l=a+(n-1) d$

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

$\Rightarrow \quad(n-1)=98 \Rightarrow n=99$

$\therefore$ Sum of terms between 100 and 200 ,

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

$\Rightarrow$ $S_{99}=\frac{99}{2}[2(101)+(99-1) 1]=\frac{99}{2}[202+98]$

$=\frac{99}{2} \times 300=99 \times 150=14850$

From Eq. (i), sum of the integers between 100 and 200 which is not divisible by 9

$=14850-1683$ [from part (i)]

$=13167$

Hence, the required sum is 13167.