Question

17. Find the sum of natural numbers between 101 and 999 which are divisible by both 2 and 5
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Answer 1

Answer:

48950

Step-by-step explanation:

Range given  = 101 to 999

divisible by both 2 & 5 => divisible by 10

a, first number = 110

l, last number  = 990

obviously d = 10

therefore n = [(l-a)/d] + 1  =  [(990-110)/10] + 1 = [880/10] + 1 = 88+1 = 89

ie,we want sum of divisors up to 89 terms

S89 = (89/2) [110+990]             by formula Sn = (n/2) * [a+l]

       =  (89/2) [1100]  = 89*550 = 48950