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sum of even integers

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by nidhis.1408 » Mon Nov 14, 2011 7:44 am
For any positive integer n , the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301?

a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150


This problem is from OG, question no- 157

I need a clearer explanation for this.
Thanks
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by GMATGuruNY » Mon Nov 14, 2011 7:53 am
nidhis.1408 wrote:For any positive integer n , the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301?

a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150


This problem is from OG, question no- 157

I need a clearer explanation for this.
Thanks
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by CappyAA » Mon Nov 14, 2011 7:54 am
I didn't use that specific formula for this question. For any evenly spaced series, the sum of the series will be equal to the average multiplied by the number of terms.

In this case, the average in a series of even numbers from 99 to 301 is 200 (our series starts with 100 and ends with 300). There are 101 terms in our series. The sum of all of these integers will be equal to 200*101 = 20,200.

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by Anurag@Gurome » Mon Nov 14, 2011 11:33 pm
A different approach using the formula given in the question:
The sum of the given even integers is 100 + 102 + 104 + ......+ 300 = 100 + (100 + 2) + (100 + 4) + (100 + 6).....(100 + 200) = 100*101 + 2(1+2+3+...100) = 10100 + 2*100*101/2 = 10100 + 10100 = 20200
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