**Question:**

Find the sum of the integers between 100 $=\frac{11}{2} \times 306$d 200 that are divisible by 9.

**Solution:**

We have to find the sum of integers between 100 and 200 that are divisible by 9.

Integers divisible by 9 between 100 and 200 are

108, 117, 126, … 198

The above equation forms an Arithmetic Progression (A.P)

Let there be n such integers in the above A.P

We know that the nth term of an A.P $a_{n}=a+(n-1) d$

Where = First term of A.P

= Common difference of successive members

Therefore the last term of above A.P (nth term of A.P) $=108+(n-1) 9$

$198=108+(n-1) 9$

$198=108+9 n-9$

$9 n=99$

$\mathrm{n}=11$

Therefore total number of integers in the A.P = 11

We know that the sum of the $n$ terms of an arithmetic progression $\mathrm{S}_{n}=\frac{n}{2}(2 a+(n-1) d)$

In the above A.P $S_{11}=\frac{11}{2}(2 \times 108+(11-1) 9)$

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

$=1683$