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The mean/median of a ten term sequence of positive consecutive even in

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The mean/median of a ten term sequence of positive consecutive even in  [#permalink]

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New post 05 Nov 2018, 23:22
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The mean/median of a ten term sequence of positive consecutive even integers is a two digit perfect square. How many such sequences are there?

A. 1
B. 2
C. 3
D. 4
E. 5
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The mean/median of a ten term sequence of positive consecutive even in  [#permalink]

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New post 06 Nov 2018, 01:16
gracie wrote:
The mean/median of a ten term sequence of positive consecutive even integers is a two digit perfect square. How many such sequences are there?

A. 1
B. 2
C. 3
D. 4
E. 5



The median/mean of a ten term sequence of positive consecutive even integers is an odd number.

So, we need to identify how many odd two digit perfect squares exist, which is \(25,49,81(5^2,7^2,9^2)\)

In order for the above numbers to be median, the 5th and 6th terms of the sequence will be as follows:
...... 24,26...... | ......48,50...... | .......80,82....... (Mean (or) Median = 5th element + 6th element/2)

Therefore, there are 3(Option C) sequences which yield a positive odd square as its median.
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Re: The mean/median of a ten term sequence of positive consecutive even in  [#permalink]

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New post 06 Nov 2018, 01:39
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gracie wrote:
The mean/median of a ten term sequence of positive consecutive even integers is a two digit perfect square. How many such sequences are there?

A. 1
B. 2
C. 3
D. 4
E. 5


Let's try to understand the question first:
Positive consecutive even integers such as 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Mean/median = 11
We need the mean to be a perfect square. Since there are 10 numbers in the list and they are even consecutive, the middle number between 5th and 6th number will be odd.

The number of two digit odd perfect squares is 5^2 = 25, 7^2 = 49, 9^2 = 81

So the numbers must be:
... 22, 24, 26, 28 ...
... 46, 48, 50, 52 ...
... 78, 80, 82, 84 ...

Answer (C)
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Re: The mean/median of a ten term sequence of positive consecutive even in  [#permalink]

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New post 05 Apr 2019, 04:11
gracie wrote:
The mean/median of a ten term sequence of positive consecutive even integers is a two digit perfect square. How many such sequences are there?

A. 1
B. 2
C. 3
D. 4
E. 5


I used this approach:

So, first I listed all the 2 digit perfect squares: 16,25,36,49,64, and 81

Now, the median=sum*n

- the sum will be even (even + even=even)

- n= 10, which is also even

So, the median= even*even=even

therefore the perfect square two digit median has to be either: 16,36, or 64--->c) 3

(I assumed these were arithmetic sequences, is that right to assume?)
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Re: The mean/median of a ten term sequence of positive consecutive even in   [#permalink] 05 Apr 2019, 04:11
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