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.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

Common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = \( \dfrac{895}{5}\)

n = 179

We know,

S_n = \( \dfrac{n}{2}[2a+(n-1)d]\)

= \( \dfrac{179}{2}[2(105)+(179-1)(5)]\)

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

= \( 179[105+(89)5]\)

= \( (179)(550)\)

= 98450