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

Question:

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

Solution:

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 .

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