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Question
Find the sum of first 100 even number which are divisible by 5.
Solution
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 – l0) = 10 and n = 100
The sum of n terms of an AP is given by
`s_n = n/2 [ 2a +(n-1) d]`
`= (100/2) xx [ 2 xx10 +(100-1) xx 10 ] [ ∵ a = 10 , d = 10 and n = 100]`
`= 50 xx[ 20+ 990]=50xx 1010 = 50500`
Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.
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