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

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Intern
Joined: 06 Oct 2010
Posts: 40
For any positive integer n, the sum of the first n positive [#permalink]

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15 Nov 2010, 16:34
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Question Stats:

67% (02:05) correct 33% (01:22) wrong based on 196 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-any-positive-integer-n-the-sum-of-the-first-n-positive-127817.html
[Reveal] Spoiler: OA
Intern
Joined: 06 Oct 2010
Posts: 40
Re: ... the sum of all the even integers between 99 and 301? [#permalink]

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15 Nov 2010, 16:38

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 *"?
Intern
Joined: 18 Aug 2009
Posts: 41
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V42
GPA: 3.29
WE: Engineering (Consulting)
Re: ... the sum of all the even integers between 99 and 301? [#permalink]

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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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Joined: 02 Jul 2009
Posts: 60
Re: ... the sum of all the even integers between 99 and 301? [#permalink]

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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

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Intern
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Re: ... the sum of all the even integers between 99 and 301? [#permalink]

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15 Nov 2010, 21:33
Awesome, chaoswithin and lotus. Thanks for your input. The problem makes a lot more sense now.
Intern
Joined: 25 Jan 2012
Posts: 4
Re: ... the sum of all the even integers between 99 and 301? [#permalink]

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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) ?
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Re: For any positive integer n, the sum of the first n positive [#permalink]

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08 Dec 2013, 18:32
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Re: For any positive integer n, the sum of the first n positive [#permalink]

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09 Dec 2013, 01:06
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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

Approach #1:
Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: totally-basic-94862.html#p730075);

The sum = 200*101= 20,200.

Approach #2:
Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-any-positive-integer-n-the-sum-of-the-first-n-positive-127817.html
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Re: For any positive integer n, the sum of the first n positive   [#permalink] 09 Dec 2013, 01:06
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