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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, 02:27
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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, 02:45
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, 02:52
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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, 12:42
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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


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, 07:41.
Last edited by HUNTdGMAT on 14 Sep 2017, 12:42, 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, 07:50
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 14 Sep 2017, 23:40
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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, 10:33
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, 12:20
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 04 Jan 2018, 23:28
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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 +  [#permalink]

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New post 03 Dec 2018, 07:03

Official Explanation:



For any collection of evenly spaced numbers, the mean and median are always equal. Because this is set with seven members, an odd number of elements, the median is the middle number, the 4th number on the list: there are three numbers on the list below it and three above it. Thus, (k + 2) is the mean as well as the median, the middle or 4th number on the list. Since there are only seven numbers, we can simply write them out:

{k – 1, k, k + 1, k + 2, k + 3, k + 4, k + 5}

The least, three lower than the median, is (k – 1).

The greatest, three higher than the median, is (k + 5).

Their product is:

product = (k – 1)(k + 5) = k2 + 4k – 5

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

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New post 03 Dec 2018, 07:06
in OE they have assumed 7 numbers as {k – 1, k, k + 1, k + 2, k + 3, k + 4, k + 5} However I think in such cases right assumption should be {k – 3, k - 2 , k - 1, k, k + 1, k + 2, k + 3} this will simplify the sum to 7K. In complex situation this will be quiet helpful.
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Re: If the average (arithmetic mean) of seven consecutive integers is k + &nbs [#permalink] 03 Dec 2018, 07:06
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