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If the average (arithmetic mean) of seven consecutive intege
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13 Jun 2018, 22:21
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. \(k^2\) – 9
B. \(k^2\) – 2k + 1
C. \(k^2\) + 4k – 12
D. \(k^2\) + 6k + 9
E. \(k^2\) + 4k – 5
A. \(k^2\) – 9
B. \(k^2\) – 2k + 1
C. \(k^2\) + 4k – 12
D. \(k^2\) + 6k + 9
E. \(k^2\) + 4k – 5
Re: If the average (arithmetic mean) of seven consecutive intege
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27 Jun 2018, 01:39
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. \(k^2\) – 9
B. \(k^2\) – 2k + 1
C. \(k^2\) + 4k – 12
D. \(k^2\) + 6k + 9
E. \(k^2\) + 4k – 5
A. \(k^2\) – 9
B. \(k^2\) – 2k + 1
C. \(k^2\) + 4k – 12
D. \(k^2\) + 6k + 9
E. \(k^2\) + 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 = \(\frac{{(K - 1) + (K) + (K + 2) + (K + 3) + (K + 4) + K + 5)}}{7}\)= k + 2
So the greatest number = k + 5 and the smallest number = K - 1
So the product =(k + 5) * (k - 1) = \(k^2\) + 4k – 5
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If the average (arithmetic mean) of seven consecutive intege
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08 Jul 2021, 09:26
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 \(4^{th}\) 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)
= \(k^2\) - k + 5k - 5
= \(k^2\) + 4k - 5
So, Answer will be E
Hope it helps!
To learn more about Statistics watch the following video
The average (arithmetic mean) of seven consecutive integers is k + 2
=> Middle number (or the \(4^{th}\) 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)
= \(k^2\) - k + 5k - 5
= \(k^2\) + 4k - 5
So, Answer will be E
Hope it helps!
To learn more about Statistics watch the following video
