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

Question

Philoid · Accepted Answer

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

Let the first term be ‘a’ and common difference ‘d’.

Let n be the total number of terms in the series.

Here, first term, a = 105

Common difference, d = 5

If l denotes the last term of the series

Then, l = a + (n – 1) × d

Here, l = 995

⇒ 995 = 105 + (n – 1) × 5

⇒ 995 – 105 = (n – 1) × 5

⇒ 890/5 = n – 1

⇒ 178 + 1 = n

∴ n = 179

Sum of A.P. =

⇒ S_n = 179 × 550 = 98,450.