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Manager
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Joined: 14 Apr 2010
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qs [#permalink] New post 23 Jul 2010, 21:40
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Question Stats:

40% (00:00) correct 60% (00:59) wrong based on 5 sessions
What is 1 + 2+ ..+98 ?
4750
4763
4790
4801
4851

1+2+..+98 = 1+(2+98)+(3+97)+...+(49+51)+50 = 51+100*48
I can't follow this explanation. Can someone provide an alternative?
Manager
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Joined: 16 Apr 2010
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Re: qs [#permalink] New post 23 Jul 2010, 21:43
Hi,

The sum of first n integers is equal to n(n+1)/2 = 98(99)/2 = 4851

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Jack
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Concentration: Marketing, General Management
WE: Business Development (Consumer Products)
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Re: qs [#permalink] New post 24 Jul 2010, 22:03
jakolik wrote:
Hi,

The sum of first n integers is equal to n(n+1)/2 = 98(99)/2 = 4851

regards,
Jack

Exactly, u just need to remember it.
SVP
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Re: qs [#permalink] New post 04 Aug 2010, 11:49
The steps to find the sum of a series of evenly spaced numbers.

(1) Find the average of the First and Last number

(1 + 98)/2 = 49.5

(2) Count the number of terms

Last - First + 1 = 98 - 1 + 1 = 98

(3) Sum = Average x Number of terms

Sum = 49.5 x 98 = 4851
Re: qs   [#permalink] 04 Aug 2010, 11:49
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