For a certain set of n numbers, where n>1, is the average

[Guest]

For a certain set of n numbers, where n>1, is the average (arithmetic mean) equal to the median?
(1) If the n number in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of the n numbers in the set is 2(n-1).

Please explain: Answer is A.

[Nov1907]

A neat little trick to remember is that for any series that is an Arithmetic progression, namely difference between each successive term is constant, the median is always = mean. I'll try to prove it below. Let us say there are n terms. There are two pssibilities n is odd or n is even. Let us say the constant difference is d (2 in the case of this problem).

1st term: a, 2nd term = a+d, 3rd term = a+2*d .... nth term = a+(n-1)*d
Adding all you get Sum = a*n +[ (1+2+3...+(n-1))]*d = a*n +[ n*(n-1)/2]*d ( another interesting result sum of the first n-1 integers is n*(n-1)/2]. Hence the Average = Sum/n = a+(n-1)/2*d

n is odd: Median = (n+1)/2 th term = a+(n-1)/2*d = Average so if n is odd we have proved avg always equal to mean.
n is even: Median = average of n/2 and (n/2+1)th term =[ a+(n/2-1)*d+a+(n/2)*d ]/2 = a+(n-1)*d/2 = Average so if n is even too we have proved avg always equal to mean. Thus (1) is always sufficient to answer such questions of if avg = median.

(2) Since only range is given we cant determine anything about the numbers in between so this informatino is insufficient. Hence answer is A.

[RonPurewal]

Yeah. As for (1) - in a little simpler terms - if the numbers in the set are EQUALLY SPACED apart (this is the meaning of 'arithmetic progression'), then the median and the mean will be equal. In fact, in any set where there is left-right SYMMETRY if the numbers are plotted on a number line, the mean and the median will be the same.

[sandeep.19+man]

I answered D for the above question. And now I think I know why B is not sufficient. Can someone please confirm?

First term = t1 = a
Last term = tn = a+(n-1)d
Range = tn-t1 = (n-1)d

(2) The range of the n numbers in the set is 2(n-1).

i.e. 2(n-1) = (n-1)d

Either n-1 = n-1 and d=2

In this case, the series become an AP and hence sufficient

OR
n-1 = 2 and d= n-1
i.e. n = 3 and d=n-1

In this case the series is a, a+(1)(2-1), a+(2)(3-1)
i.e. a, a+1, a+4 which is not an AP and hence mean <> median

[gokul_nair1984]

I answered D for the above question. And now I think I know why B is not sufficient. Can someone please confirm?

First term = t1 = a
Last term = tn = a+(n-1)d
Range = tn-t1 = (n-1)d

(2) The range of the n numbers in the set is 2(n-1).

i.e. 2(n-1) = (n-1)d

Either n-1 = n-1 and d=2

In this case, the series become an AP and hence sufficient
OR
n-1 = 2 and d= n-1
i.e. n = 3 and d=n-1

In this case the series is a, a+(1)(2-1), a+(2)(3-1)
i.e. a, a+1, a+4 which is not an AP and hence mean <> median

Sandeep you are right, but plugging in numbers would be much simpler in this case. Let me show you how:

We are given range,R =2(n-1).

** No point checking for n=2 because it will always satisfy the condition as the average of 2 numbers is always the central number(median) for all 2 digit numbers.

Let n=3 =>R=4

The number set can be: {2,___,6} (Because the range has to be 4.)

If the central number is 4, then the median (4) will be equal to avg(4). What if the central number is 3. Then the median(3) will not be equal to the average (11/3). Hence (B) is insufficient.You can also consider decimal values as the middle number to disprove (B)...

[tim]

thanks, Gokul..

[rajivmgmat]

I am sorry I could not follow Gokul's explanation, which is pretty much on the same lines as the OG-Quant explanation. If d can be proved to be 2 then there is no way the central number could be 3, unless the 1st number is odd (ie. 1)
Sandeep's analysis seems flawed also, in my view.
Taking Sandeep's cue...Once we get the equation
(n-1)d = 2(n-1)
It is given that n>1, which means n<>1, therefore we can divide both sides by (n-1), which gives d=2
If d=2 it provides me the same information as does A, does it not?
In which case how is B insufficient?

Even if we follow Sandeep's 2nd case ie.
if (n-1)d=2(n-1) then
he says that it could be that d=n-1 and n-1=2 ie. n=3
But this gives us the same result if we substitute n=3 in d=n-1, ie. d=2
Either way d=2, therefore its an arithmetic progression.

It seems D is the correct answer. Please do explain where I got it wrong.

[gokul_nair1984]

Rajiv,I have not clearly understood your doubt. Let me rephrase..

I hope you have understood how A is sufficient.

Coming to B,

The range of the n numbers in the set is 2(n-1).---Given
You can ignore the case where n=2 because if you take only 2 numbers, the mean will always be the median.eg(3,5)..
(-100,80)...et al.... I hope we are on the same page until now.


If n=3, R=4. Let's fix the first number in this set to 2(just for simplicity)

{2,_x__,_y__}. If you need the Range to be 4, then y-2=4 =>y=6

So, {2,x,6}---R=4

Case 1:Let x=4, then median=4 and
mean=(2+4+6)/3=4
Hence Mean=Median

Case 2: Let x=5 , so the set becomes {2,5,6}--Here mean=13/3=4.3 and median =5.
Hence Mean!=Median.

So we have the answer to the question stem as Yes in one case and No in the other, which makes this statement invalid.

IS it clear?

[rajivmgmat]

Thanks Gokul, for attempting clarify. I am not sure I fully subscribe to your line of thought, but your examples have given me the answer, I think.

See, All (2) states is that the range of the n numbers in the set is 2(n-1). Nowhere does it say that the numbers are in arithmetic progression.
Therefore the formula
Sn-a = (a+(n-1)d) -a cannot be applied

With this given, your examples {2,4,6} and {2,3,6} can be applied. Both validate the given condition of range= 2(n-1) but the latter gives different mean and median. Therefore (2) is indeed insufficient...and both you and GMAC are correct :)

Thanks again, Gokul.

[RonPurewal]

this problem turns on one fact that is definitely worthwhile for all of you to memorize:
in ANY set of numbers that are equally spaced (i.e., an "arithmetic sequence", if you happen to know that terminology), the MEAN and the MEDIAN are the same.

if you realize this fact, then the sufficiency of statement (1) is immediate.

--

if you don't make the realization above, just throw in a whole bunch of random sequences that satisfy statement (1), and just check whether they answer the question in a consistent way. in other words, just throw in a bunch of sequences with a consistent difference of 2.

try {2, 4} --> mean = median = 3
try {11, 13, 15, 17, 19} --> mean = median = 15
etc.
you should try a few examples (without taking a ridiculous amount of time), just to be fully convinced, but you will discover in each case that the mean and median are the same value. so statement 1 is sufficient.

--

statement 2 isn't sufficient because the range has only to do with the largest and smallest values in the set; you can tweak the values in the middle of the set in ways that will change the mean and median.
for instance, in the above-mentioned set {11, 13, 15, 17, 19}, the mean and the median are both 15.
if you change just the second value from 13 to, say, 12, then the median will still be 15 -- but the mean will now be something less than 15. (no need to calculate it; the mere fact that it has decreased is enough to conclude that it is no longer equal to 15.)

--

answer (a).