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

Question

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

KarishmaB · Accepted Answer

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)

pushpitkc · Answer

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.

Ish1996 · Answer

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?)

chetnasharma · Answer

Hi, I'm not understanding the odd-even approach here. Can someone please explain again? Thank you.

Posted from my mobile device

HarshavardhanR · Answer

- We are dealing with a sequence of 10 terms of positive consecutive even integers. Example: 2,4,6,8,10,12,14,16,18,20.
- The mean/median of such a set = (10th term + 11th term)/2. Example: Mean or Median = (10+12)/2 = 11.

- Let's understand this better.

Say, 
10th term = a
Then,
11th term = a+2 (next even integer)

Mean or Median = (a + a+2) / 2 = a+1.

Observe:  By default, given that a is an even number, the mean/median,  a+1 is an ODD number .

Now,
(1) Mean/Median, a+1, is an odd number.
(2) Mean/Median, a+1, is also a 2-digit perfect square.

So, we need to see what are the various 2-digit odd perfect squares.

25, 49, 81. This is all.

So, we should have 3 sequences of consecutive even positive integers in total where the mean or median is a 2-digit perfect square.  Choice C.

(1) ....22, 24,26 ,28....
(2) ....46, 48,50 ,52.....
(3) ....78, 80,82 ,84.....

Hope this helps.
___
Harsha
Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

Stella5200 · Answer

Median term for even number of terms is the midpoint of the n/2 and (n/2) +1 th terms. here its the 5th and 6th.

The midpoint of two consecutive even terms is odd (10,12) = 11, (12,14) = 13.

Even if you dont use that. use sequences:

5th term is a+4d 
6th is a+5d. 
d here is 2

so 2a + 9(2) = 2a + 18 and then div by 2 to get the median

you get a + 9.

remember a is the first term and the questions says all terms are even. so E + O = O. which means median is odd.

So in the list of perfect 2 digit squares there are only 3 odd.

3 is the answer.

Adit_ · Answer

All the 2-digit odd perfect squares form a squence = 5^2,7^2,9^2. Therefore  3 .

____________________________________

I think question would have been more tricky if the rule was perfect squares>5 and <100. Because that adds the layer of knowing that the mean is an odd number. This way from this question even of odd one can arrive at 3.

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

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