Find the sum of all multiples of 9 lying between 300 and 700.

The multiples of 9 lying between 300 and 700 are 306, 315, ..., 693.

This is an AP with a = 306, d = 9 and l = 693.

Suppose there are n terms in the AP. Then,

$a_{n}=693$

$\Rightarrow 306+(n-1) \times 9=693 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 9 n+297=693$

$\Rightarrow 9 n=693-297=396$

$\Rightarrow n=44$

$\therefore$ Required sum $=\frac{44}{2}(306+693) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

$=22 \times 999$

$=21978$

Hence, the required sum is 21978.