Question

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

Answer 1

Explanation:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995. This sequence forms an A.P. Here, first term, a = 105 Common difference, d = 5 Here, a+(n−1)d = 995 => 105+(n−1)5 = 995 => (n−1)5 = 995−105 = 890 => n−1 = 178 => n = 179 Sn = n/2[2a+(n−1)d] ∴ Sn = 179/2[2×(105)+(179−1)×(5)] = 179/2[2(105)+(178)(5)] = 179[105+(89)5] = (179)[105+445] = 179×550 = 98450Read more on Sarthaks.com - https://www.sarthaks.com/1553/find-the-sum-of-all-natural-numbers-lying-between-100-and-1000-which-are-multiples-of-5

Answer 2

a = 105

l = 995 = a+(n-1)d
995 = 105+(n-1)5
890 = (n-1)5 ————> 178+1=179=n

d = 5

Sum = n(a+l)/2
= 179(105+995)/2 = 179*1100/2 = 98450

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