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# qs

Author Message
Manager
Joined: 14 Apr 2010
Posts: 218

Kudos [?]: 235 [0], given: 1

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23 Jul 2010, 22:40
00:00

Difficulty:

(N/A)

Question Stats:

40% (00:00) correct 60% (00:59) wrong based on 5 sessions

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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?

Kudos [?]: 235 [0], given: 1

Manager
Joined: 16 Apr 2010
Posts: 214

Kudos [?]: 142 [0], given: 12

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23 Jul 2010, 22:43
Hi,

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

regards,
Jack

Kudos [?]: 142 [0], given: 12

Manager
Joined: 03 Jun 2010
Posts: 172

Kudos [?]: 71 [0], given: 40

Location: United States (MI)
Concentration: Marketing, General Management

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24 Jul 2010, 23:03
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.

Kudos [?]: 71 [0], given: 40

VP
Joined: 17 Feb 2010
Posts: 1476

Kudos [?]: 759 [0], given: 6

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04 Aug 2010, 12:49
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

Kudos [?]: 759 [0], given: 6

Re: qs   [#permalink] 04 Aug 2010, 12:49
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