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