Question

Find the sum of all natural numbers lying between 100 and 500 which are divisible by 8

Answer 1

a.1 = 104(this is the first number divisible by 8 after 100)

d = 8

a.n = 496(this is last number divisible by 8 before 500)

a.1 + (n-1) d = 496

104 + (n-1) 8 = 496

(n-1) 8 = 496 - 104

(n-1) 8 = 392

n - 1 = 392/8

n - 1 = 49

n = 50

S.n = n/2 [2a + (n-1) d]

S.50 = 50/2 [2 x 104 + (50-1) 8]

S.50 = 25 [208 + (49) 8]

S.50 = 25 [208 + 392]

S.50 = 25 [600]

S.50 = 15000

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