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For any positive integer n, the sum of the first n positive

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For any positive integer n, the sum of the first n positive [#permalink] New post 15 Nov 2010, 16:34
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Question Stats:

70% (01:46) correct 30% (01:09) wrong based on 20 sessions
For any positive integer n, the sum of the first n positive integers equals \frac{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
[Reveal] Spoiler: OA
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Re: ... the sum of all the even integers between 99 and 301? [#permalink] New post 15 Nov 2010, 16:38
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The thing I don't get is why 2 * \frac{150(150+1)}{2} is equal to the sum of the first 150 even
integers. Why the "2 *"?
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Intern
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Re: ... the sum of all the even integers between 99 and 301? [#permalink] New post 15 Nov 2010, 17:09
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Sum of the first 150 positive integer is (1 + 2 + 3 + 4 + ... + 150)

multiply that by 2 and you get (2 + 4 + 6 + 8 + ... + 300) which is the sum of first 150 positive even integers.
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Re: ... the sum of all the even integers between 99 and 301? [#permalink] New post 15 Nov 2010, 17:40
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I would solve it in a different way.

First of all, total number of even integers between 99 and 301 are, (301-99)/2 = 202/2=101

Average = (301+99)/2 = 400/2 = 200

Sum = Average*total count = 200*101 = 20,200

Answer is B
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Re: ... the sum of all the even integers between 99 and 301? [#permalink] New post 15 Nov 2010, 21:33
Awesome, chaoswithin and lotus. Thanks for your input. The problem makes a lot more sense now.
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Re: ... the sum of all the even integers between 99 and 301? [#permalink] New post 10 Mar 2012, 10:41
Hi,

I understand how Lotus has solved this problem. But I am getting confused.

If I use the formula that the question asks us to use, then wouldnt it be

[(101) (102)]/2? where n =101 (the number of even integers in this case) ?
Re: ... the sum of all the even integers between 99 and 301?   [#permalink] 10 Mar 2012, 10:41
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