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The sum of all the integers k such that –26 < k < 24 is

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The sum of all the integers k such that –26 < k < 24 is [#permalink]

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New post 27 Aug 2010, 23:09
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The sum of all the integers k such that –26 < k < 24 is

A. 0
B. -2
C. -25
D. -49
E. -51

What is the trap here...I am getting answer as -51 where as OA is diff.
[Reveal] Spoiler: OA
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Re: The sum of all the integers k such that –26 < k < 24 is [#permalink]

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New post 28 Aug 2010, 07:11
-25------------------0-----------------23

values upto + 23 cancels out & we are left with only -25 & -24 sum of which is - 49. hence option D.

Since this is in equality we dont consider -26 and 24 as values of K. K is greater than -26 & less than +24 therefore range of K is -25 to 23. You might be considering -26 & 24, in that case the range becomes -26 to +24 and the answer will be 51 which is wrong.
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The sum of all the integers k such that –26 < k < 24 is [#permalink]

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New post 28 Aug 2010, 07:21
udaymathapati wrote:
The sum of all the integers k such that –26 < k < 24 is
A. 0
B. -2
C. -25
D. -49
E. -51

What is the trap here...I am getting answer as -51 where as OA is diff.


Notice that the range is not inclusive: k can not be -26 or 24! That's why you are getting -51.

The sum will be: (-25)+(-24)+(-23)+...+(-1)+0+1+..+23 --> the sum of pairs -23 and 23, -22 and 22 and so on is 0 and we are left only with -25+(-24)=-49.

Or: as we have evenly spaced set: the sum will be average of the first and the last terms multiplied be the # of terms: \(\frac{-25+23}{2}*49=-49\).

Answer: D.

Hope it's clear.
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The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 03 Jan 2012, 09:16
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The sum of all the integers k such that -26 < k < 24 is

A. 0
B. -2
C. -25
D. -49
E. -51

[Reveal] Spoiler:
Whether the following approach correct?
K takes value form -25 to 23
Thus number of values (n) 23+25+1 = 49
Now, n(a1+an)/2
49(-25+23)/2 = -49

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Last edited by Bunuel on 04 Sep 2015, 05:33, edited 2 times in total.
Renamed the topic and edited the question.
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Re: The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 03 Jan 2012, 09:58
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Baten80 wrote:
The sum of all the integers k such that -26<k<24 is
A. 0
B. -2
C. -25
D. -49
E. -51


Easy one -
-25, -24, -23,-22,...... -1,0, 1, 2...., 22, 23
Cancel everyhitng and we're left with - -25 and -24 = -49.
D is the answer.

Cheers!
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Re: The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 03 Jan 2012, 11:10
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or: if we have enough time, we can use the following formula, since its an AP:
sum=n/2*{2a+(n-1)*d}
a= -25, d=1, n=49 (including 0)
therefore, sum= 49/2(-50+48)
=-49
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Re: The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 04 Jan 2012, 13:05
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Sum = mean X number of elements
mean = [23 + (-24)]/2 = -1
number = [23 - (-24)]/1 + 1 = 49
Sum = -1 x 49 = -49
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Re: The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 04 Jan 2012, 22:18
C

Mean = (-25 + 23)/2 = -1
Number count = 25 + 23 + 1 = 49
Sum = Mean * Count = -49
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Re: The sum of all the integers k such that -26 < k < 24 is [#permalink]

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New post 04 Sep 2015, 02:09
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Re: The sum of all the integers k such that -26 < k < 24 is   [#permalink] 04 Sep 2015, 02:09
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