Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

Here, $a=105$ and $d=5$

$a+(n-1) d=995$

$\Rightarrow 105+(n-1) 5=995$

$\Rightarrow(n-1) 5=995-105=890$

$\Rightarrow 105+(n-1) 5=995$

$\Rightarrow n-1=178$

$\Rightarrow n=179$

$\therefore S_{n}=\frac{179}{2}[2(105)+(179-1)(5)]$

$=\frac{179}{2}[2(105)+(178)(5)]$

$=179[105+(89) 5]$

$=(179)(550)$

$=(179)(105+445)$

$=98450$

Thus, the sum of all natural numbers lying between 100 and 1000 , which are multiples of 5 , is 98450 .