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Consecutive Integers question - Repeated

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Consecutive Integers question - Repeated [#permalink] New post 20 Apr 2010, 05:16
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Difficulty:

  5% (low)

Question Stats:

42% (01:36) correct 57% (00:57) wrong based on 7 sessions
Hi Guys - can someone please let me know the concept behind this question and how to solve.

For any positive integer n, the sum of the first n positive integres equals n(n+1)/2. What is the sum of all even integers between 99 & 301?

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

I understand the concept behind n(n+1)/2 and also the sum the question is asking for i.e sum of even integers between 100 & 300.

Can someone please explain the details around how to solve this?
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Re: Consecutive Integers question - Repeated [#permalink] New post 20 Apr 2010, 06:49
gsaxena26 wrote:
Hi Guys - can someone please let me know the concept behind this question and how to solve.

For any positive integer n, the sum of the first n positive integres equals n(n+1)/2. What is the sum of all even integers between 99 & 301?

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

I understand the concept behind n(n+1)/2 and also the sum the question is asking for i.e sum of even integers between 100 & 300.

Can someone please explain the details around how to solve this?


In a set of consecutive integers the sum is equal to the number of terms * the average (or median because in consecutive terms mean = median)

Average = 301 + 99 / 2 = 200

Number of terms: 100 is the 50th even term and 300 is the 150th even term. (2 is the first even term and 100/2 = 50 or the 50th term)

150-50 + 1 = 101

Answer b: 20,200
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Re: Consecutive Integers question - Repeated [#permalink] New post 21 Apr 2010, 03:08
Thanks lagomez. Really appreciate your help. :lol:
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Re: Consecutive Integers question - Repeated [#permalink] New post 20 Mar 2012, 05:43
answer is D: 40,200

first term and last term between 99 and 301 are respectively 100 and 300

and the number of terms is: 300-100+1 which is 201

thus the answer equals (100+300)/2*201=40,200
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Re: Consecutive Integers question - Repeated [#permalink] New post 21 Mar 2012, 22:49
20200...that's option B
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Re: Consecutive Integers question - Repeated [#permalink] New post 22 Aug 2012, 08:38
its 20200

took a bit long time ..
analysing extra bit of useless info
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Re: Consecutive Integers question - Repeated   [#permalink] 22 Aug 2012, 08:38
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