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?
Answer Please!!!!
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- f2001290
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Add -25 + -24 + -23 + .... + 0 + 1 + 2 +3 + ... -23Shadow 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?
We are left out with -25 + -24 = -49
Hence D
- beeparoo
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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
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