SOLUTION: If the average (arithmetic mean) of five consecutive integers is m, what is the product of the greatest and least of these five integers?
a) 4 b) (m+2)^2 c) 2m d) 4m^2 e) m^2 -4
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a) 4 b) (m+2)^2 c) 2m d) 4m^2 e) m^2 -4
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Question 238964: If the average (arithmetic mean) of five consecutive integers is m, what is the product of the greatest and least of these five integers?
a) 4 b) (m+2)^2 c) 2m d) 4m^2 e) m^2 -4
You can put this solution on YOUR website! If the average (arithmetic mean) of five consecutive integers is m, what is the product of the greatest and least of these five integers?
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Let the consecutive integers be:
m-2
m-1
m
m+1
m+2
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The sum is 5m
The average is m
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Equation:
Product of greatest and least is (m-2)(m+2) = m^2-4
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Cheers,
Stan H.
a) 4 b) (m+2)^2 c) 2m d) 4m^2 e) m^2 -4
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