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If the average (arithmetic mean) of seven consecutive integers is k +

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If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 13 Sep 2017, 03:27
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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 - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

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Re: If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 13 Sep 2017, 03:45
Let first integer of series = a
Difference between integers = d= 1 (consecutive integers given)
Here no. of terms in series = n=7
Average = K+2


Average = Arithmetic mean =\(\frac{sum of all 7 numbers}{7}\)

by formula Sum of series =\(\frac{n(2a+(n-1)d)}{2} = \frac{7(2a+(7-1)*1)}{2} = \frac{7(2a+6)}{2} = 7(a+3)\)

As Average = k+2 = \(\frac{sum of all 7 numbers}{7}\) = \(\frac{7(a+3)}{7}\)
=> k+2 =a+3
=> a= k-1

Now as nth term of series = a+(n-1)d => last term of given series = a+(7-1)*1 = (k-1)+6 = k+5

Product of first and last term = (k-1)*(k+5) = k^2+5k-k-5 = k^2+4k-5

Answer: C
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Re: If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 13 Sep 2017, 03:52
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Bunuel 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 - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12


Let the first integer be x;
Sum of the 7 integers x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)= 7x+21

Average = Sum/n= 7(x+3)/7=x+3;

x+3= k+2 ==>x= k-1;
Greatest integer = k+5;

Product x*x+6 =k^2+4k-5

Ans:C
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If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post Updated on: 14 Sep 2017, 13:42
Bunuel 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 - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12


Since there are 7 consecutive integers, K+2 is the fourth number in the series. So, the series should be k-1, k, k+1, k+2, k+3, k+4, k+5.

then the product of the greatest and least integer =(k+5)* (k-1)= k^2+4k-5

Ans is C
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Originally posted by HUNTdGMAT on 13 Sep 2017, 08:41.
Last edited by HUNTdGMAT on 14 Sep 2017, 13:42, edited 1 time in total.
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If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 13 Sep 2017, 08:50
Given data: A set with 7 consecutive numbers has mean k+2

Lets assume the 7 numbers to be 1,2,3,4,5,6,7
The mean of consecutive integers in a set with odd number of elements is the middle number.
In this set, the 4th element which is 4 is the mean.
k+2 = 4 => k = 2
The product of the least and the greatest number is 1*7(7)

Evaluating the answer options(s.t k = 2)
A. k^2 - 9 = 4 -9 = -5
B. k^2 - 2k + 1 = 4 - 4 + 1 = 1
C. k^2 + 4k - 5 = 4 + 8 - 5 = 7

Since Option C gives us the same value for the product of the smallest and greatest number,
we need not evaluate the other 2 answer options. (Option C) is our answer!
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If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 15 Sep 2017, 00:40
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Total 7 consecutive nos and K+2 is the average which means 3 nos each are on the right and left of this no.
K-1, k,k+1,k+2,k+3, k+4, k+5
Smallest no=k-1
largest no=k+5
Product=(k-1)(k+5)=k²+4k-5=C is the answer
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If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 15 Sep 2017, 11:33
Bunuel 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 - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

I used algebra, erred, and took a long time to correct. So I timed myself using what I thought was a time-consuming method. Wrong. I was well under a minute.

I used only one sketch, and no fancy lines or asterisks. I used them here for clarity.

1. List the integers

x|x+1|x+2|x+3|x+4|x+5|x+6|

2. Place (k+2) under the median

"The arithmetic mean is k + 2." Median = mean for evenly spaced set. So

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|***|***|k+2|***|***|***|

3. Find k. From above:
k + 2 = x + 3
k = x + 1, so

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|_k_|***|k+2|***|***|***|

4. Find answer. The product of the greatest and least integer is? We need x, and (x+6), in terms of k, so

_x_|x+1|x+2|x+3|x+4|x+5|x+6|
k-1|_k_ |***|k+2|***|***|k+5|

(k - 1)(k + 5) = k\(^2\) + 4k - 5

ANSWER C
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Re: If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 13 Oct 2017, 13:20
Bunuel 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 - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12


Is this a 500 lvl question? really doesn't feel like it

anyway

Sum of the numbers = 7k + 14
let x = smallest number
21 + x = 7k + 14
x= k-1

product of smallest and biggest = x(x+6) = x^2 + 6x = k^2 - 2k + 1 + 6k - 6 = k^2 +4k - 5

Thus, answer C

Not a hard problem, but I don't think it is level 500.
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Re: If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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New post 05 Jan 2018, 00:28
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the source is from Magoosh
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Re: If the average (arithmetic mean) of seven consecutive integers is k +  [#permalink]

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Re: If the average (arithmetic mean) of seven consecutive integers is k + &nbs [#permalink] 05 Jan 2018, 01:05
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