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prateekflex said – Mon, 20 Apr 2009 05:45:10 -0000 ( Link )
It will be option 2) 7550
Number of even integers from 1 to 100 are same as from 101 to 200. So we just need to add 100(50) to the provided total i.e 2500+100(50)=7500.
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indianschoolofbusiness said – Mon, 20 Apr 2009 09:55:32 -0000 ( Link )
The total number of numbers are 50 (102 to 200). T(n) = a + (n-1) d. => 200 = 102 + (n-1) 2. => n = 50. Therefore, sum of a series is S(n) = n/2 (a+l) [a – first no. and l – last number]. =>50/2(102+200) = 7550.
I hope this is correct.
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Sureshbala said – Tue, 21 Apr 2009 08:48:55 -0000 ( Link )
As prateekflex said, you can make use of the sum of the first 50 even integers given.
Anyway, this can still be answered quickly without that data.
Total number of even integers from 102 to 200 = 50.
Since the series is in A.P, sum = 50/2( first term + last term) = 50/2(102+200) = 50(151) =7550
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