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In a certain sequence, the term [m]A_n[/m] is given

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In a certain sequence, the term [m]A_n[/m] is given  [#permalink]

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New post 28 Jan 2019, 20:37
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A
B
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E

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Question Stats:

64% (02:51) correct 36% (03:17) wrong based on 11 sessions

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In a certain sequence, the term \(A_n\) is given by the formula \(A_n = 16 - \frac{(4-n)^3}{|n-4|}\) for all positive integers n≠4, While \(A_4 = 16\). For which integer value of k greater than 2 is the mean of all values in the sequence form \(A_1\) through \(A_k\) equal to the median of those values?

a) 6
b) 7
c) 8
d) 9
e) 10

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Re: In a certain sequence, the term [m]A_n[/m] is given  [#permalink]

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New post 28 Jan 2019, 21:06
A1=16-3
A2=16-4
A3=16-1
A4=16+/-0
A5=16+1
A6=16+4
A7=16+3

The Median and the Mean of A1-A7 are both 16. k=7→(B)

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Re: In a certain sequence, the term [m]A_n[/m] is given  [#permalink]

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New post 29 Jan 2019, 18:19
Afc0892 wrote:
In a certain sequence, the term \(A_n\) is given by the formula \(A_n = 16 - \frac{(4-n)^3}{|n-4|}\) for all positive integers n≠4, While \(A_4 = 16\). For which integer value of k greater than 2 is the mean of all values in the sequence form \(A_1\) through \(A_k\) equal to the median of those values?

a) 6
b) 7
c) 8
d) 9
e) 10


Hello Afc0892 !

Could you please provide the explanation?

Thank you in advance!
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Re: In a certain sequence, the term [m]A_n[/m] is given  [#permalink]

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New post 18 Mar 2019, 05:22
jinghao wrote:
A1=16-3
A2=16-4
A3=16-1
A4=16+/-0
A5=16+1
A6=16+4
A7=16+3

The Median and the Mean of A1-A7 are both 16. k=7→(B)

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Hi jinghao
Your answer is correct, but maybe small typo.
A1= 7
A2=12
A3=15
A4=16
A5=17
A6=20
A7=25

Mean = \(\frac{7+12+15.....25}{7}\)=16
Median =16
hence B
jfranciscocuencag, let me know if anything is still unclear.
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Re: In a certain sequence, the term [m]A_n[/m] is given   [#permalink] 18 Mar 2019, 05:22
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