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If the average (arithmetic mean) of five consecutive negative integers

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If the average (arithmetic mean) of five consecutive negative integers  [#permalink]

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New post 17 Oct 2018, 01:46
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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

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If the average (arithmetic mean) of five consecutive negative integers  [#permalink]

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New post 17 Oct 2018, 02:38
\(\frac{sum of numbers}{number of numbers}\) =average, or \(\frac{sum of observations}{n}\)=arithmetic mean
We don't know what those numbers are so 'n'. The numbers are consecutive so we can minus one each time.
\(\frac{n+(n-1)+(n-2)+(n-3)+(n-4)}{5}\) = 2k-1
\(\frac{5n-10}{5}\)=2k-1
n-2=2k-1
n=2k+1

Sub in 2k+1 for the numbers above
2k+1, 2k, 2k-1, 2k-2, 2k-3

Difference is subtraction. Doesn't matter if they are positive or negative numbers, difference is still the same
(e.g. -5--2=-3=3, or if absolutes are familiar |-5|-|-2|=5-2=3
|2k+1|-|2k-3|
2k+3-2k+1 = 4

OR

5 Consecutive integers
5, 4, 3, 2, 1
5-1 = 4
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Re: If the average (arithmetic mean) of five consecutive negative integers  [#permalink]

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New post 21 Oct 2018, 02:35
For consecutive integers, the mean 2k - 1 is also going to be the median value.

Therefore, the first value is 2k - 3 and the fifth values is 2k + 1. Difference (Range) = 2k + 1 - 2k + 3 = 1 + 3 = 4

Answer is A
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Re: If the average (arithmetic mean) of five consecutive negative integers   [#permalink] 21 Oct 2018, 02:35
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If the average (arithmetic mean) of five consecutive negative integers

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