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11 Feb 2022, 03:45
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If the average (arithmetic mean) of five consecutive negative integers is 2k – 1, what is the difference between the greatest and least of the five integers?
A. 4
B. 4k
C. 4k + 4
D. 4 - 4k
E. 4k^2 − 4k
A. 4
B. 4k
C. 4k + 4
D. 4 - 4k
E. 4k^2 − 4k
11 Feb 2022, 07:32
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Bunuel
If the average (arithmetic mean) of five consecutive negative integers is 2k – 1, what is the difference between the greatest and least of the five integers?
A. 4
B. 4k
C. 4k + 4
D. 4 - 4k
E. 4k^2 − 4k
A. 4
B. 4k
C. 4k + 4
D. 4 - 4k
E. 4k^2 − 4k
Lots of superfluous information here. In fact, it makes no difference that the integers are negative, and it makes no difference that the average of the numbers is 2k - 1.
For ANY 5 consecutive integers (positive or negative), the difference between the greatest and least value is always 4. Here's why:
Let x = the smallest integer
So, x + 1 = the next integer
x + 2 = the next integer
x + 3 = the next integer
x + 4 = the greatest integer
So, the difference between the greatest and least value = (x + 4) - x = 4
Answer: A
PRI23YA
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13 Feb 2022, 20:36
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a,b,c,d,e
Given that c=2k-1 and the integers are consecutive => diff between each successive term is 1
the series now becomes: 2k-3, 2k-2,2k-1,2k,2k+1
2k+1-(2k-3)= 4
Given that c=2k-1 and the integers are consecutive => diff between each successive term is 1
the series now becomes: 2k-3, 2k-2,2k-1,2k,2k+1
2k+1-(2k-3)= 4
