Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒100 = 2 + (n –1) 2
⇒ n = 50
The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This forms an A.P. with both the first term and common difference equal to 5.
∴100 = 5 + (n –1) 5
⇒ 5n = 100
⇒ n = 20
The integers, which are divisible by both 2 and 5, are 10, 20, … 100.
This also forms an A.P. with both the first term and common difference equal to 10.
∴100 = 10 + (n –1) (10)
⇒ 100 = 10n
⇒ n = 10
∴Required sum = 2550 + 1050 – 550 = 3050
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.
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10 Comment(s) on this Question
Thanks for the solution.
That is correct but I have asked other answer
Great..
Thanks sir it helped me
thanks sir.. ... wonderful answer.....
Thanks a lot
Thanks sir
Can it be solved by n(n+2)/2
Thanks
Not best
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