Answer Please!!!!

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Answer Please!!!!

by Shadow » Wed Jun 27, 2007 4:30 am
The sum of all the integers k such that –26 < k < 24 is

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


Could anyone please let me know the answer for the above question?

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by harry_x1 » Wed Jun 27, 2007 7:19 am
as per the statement k is a set of following integers[-25,-24,-23......,0,......22,23]
now sum of all the integers in the following set will be equal to -25+(-24)= -49 ans

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Re: Answer Please!!!!

by f2001290 » Thu Jun 28, 2007 6:04 am
Shadow wrote:The sum of all the integers k such that –26 < k < 24 is

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


Could anyone please let me know the answer for the above question?
Add -25 + -24 + -23 + .... + 0 + 1 + 2 +3 + ... -23

We are left out with -25 + -24 = -49
Hence D

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by beeparoo » Sat Jul 07, 2007 12:54 pm
Another way to look at this problem:

For CONSECUTIVE numbers in a set only...

SUM = (# of terms) x (middle number)

Determine # of terms:
23 - (-25) + 1 = 49
Thus, 49 terms in the set

Determine middle number:
[-25 + 23]/2 = -1
Thus, the middle number of the set is -1
NOTE: You can use upper and lower limit to determine the median when you have a consecutive set only

Finally, SUM = (49) x (-1) = -49

The previous methods are a lot faster and easier here. But in other questions where consecutive sets are apparent, these skills are crucial.

Later beh-beh

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by Shadow » Tue Jul 10, 2007 3:18 am
Thanks a lot for the detailed explanation which will help me solve problems of this nature