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

Asked by Pragya Singh | 1 year ago |  75

##### Solution :-

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,

Sn = $$\dfrac{n}{2}[2a+(n-1)d]$$

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

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

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

$$(179)(550)$$

= 98450

Answered by Abhisek | 1 year ago

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