Q. 24 B4.8( 4 Votes )

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

The numbers lying between 100 and 1000 that are multiples of 5 are

105, 110, 115, 120,…, 995

a2 – a1 = 110 – 105 = 5

a3 – a2 = 115 – 110 = 5

a3 – a2 = a2 – a1 = 5

Therefore, the series is in AP

Here, a = 105, d = 5 and an = 995

We know that,

an = a + (n – 1)d

995 = 105 + (n – 1)5

995 – 105 = (n – 1)5

890 = (n – 1)5

178 = (n – 1)

n = 179

Now, we have to find the sum of this AP

S179 = 179[105 + 89 × 5]

S179 = 179 [550]

S179 = 98450

Hence, the sum of all numbers lying between 100 and 1000 that are multiples of 5 is 98450.

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