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If the average (arithmetic mean) of seven consecutive intege

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amorphous
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If the average (arithmetic mean) of seven consecutive intege [#permalink] New post   13 Jun 2018, 22:21
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If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is ?

A. k2 – 9


B. k2 – 2k + 1


C. k2 + 4k – 12


D. k2 + 6k + 9


E. k2 + 4k – 5
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E
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Re: If the average (arithmetic mean) of seven consecutive intege [#permalink] New post   27 Jun 2018, 01:39
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amorphous wrote:
If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is ?

A. k2 – 9


B. k2 – 2k + 1


C. k2 + 4k – 12


D. k2 + 6k + 9


E. k2 + 4k – 5


Here
we know k + 2 is the Arithmetic Mean and there are 7 nos

SO the numbers are; K - 1, K , K + 2 , K + 3, K + 4, K + 5

i.e. arithmetic mean = (K−1)+(K)+(K+2)+(K+3)+(K+4)+K+5)7= k + 2

So the greatest number = k + 5 and the smallest number = K - 1

So the product =(k + 5) * (k - 1) = k2 + 4k – 5
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If the average (arithmetic mean) of seven consecutive intege [#permalink] New post   08 Jul 2021, 09:26
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Theory: In case of evenly spaced sets the mean of the set is the middle number (in case of odd number of numbers)

The average (arithmetic mean) of seven consecutive integers is k + 2
=> Middle number (or the 4th number) = k+2

So, the consecutive numbers less than k+2 are
k-1, k, k+1

The consecutive numbers greater than k+2 are
k+3, k+4, k+5

The product of the greatest and least integer
= (k-1) *(k+5)
= k2 - k + 5k - 5
= k2 + 4k - 5

So, Answer will be E
Hope it helps!

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If the average (arithmetic mean) of seven consecutive intege [#permalink]
08 Jul 2021, 09:26
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