Question

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

Answer 1

Natural numbers between 100 and 1000 which are multiples of 5 are 105, 110, 115, 120, ........,995.

Here a = 105, d = 110 - 105 = 5, $a_n$ = 995

We know that $a_n=a+(n-1)$

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

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

$\Rightarrow \frac{890}{5}=n-1$

$\Rightarrow n=178+1=179$

Now $s_n=\frac{n}{2}[a+l]$

$\therefore s_{179}=\frac{179}{2}[105+995]$

= $\frac{179}{2}\times 1100=98450.$