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