3
$\begingroup$

For the data set the geometric mean is 10 then arithmetic mean will be? I tried hard to calculate the arithmetic mean from geometric mean which is given as 10 but unable to find it.

$\endgroup$
  • 4
    $\begingroup$ Without some more information the question cannot be answered! Hint: What happens if $n=1$? Then make some examples with $n=2$. $\endgroup$ – kjetil b halvorsen Dec 27 '18 at 13:28
  • 2
    $\begingroup$ Assuming all numbers are positive (which is really the only situation in which computing a GM makes sense), the AM-GM inequality implies the AM is greater than or equal to 10. That is all you can possibly know from the information given. $\endgroup$ – whuber Dec 27 '18 at 13:48
3
$\begingroup$

The arithmetic mean can be any number $10$ or larger.


Let the dataset consist of the numbers

$$x_1 \ge x_2 \ge \cdots \ge x_n \gt 0$$

where $n \ge 2.$ If we write $\bar x$ for its arithmetic mean then the sum of all the numbers is just $n$ times the mean:

$$S = n \bar x = x_1 + x_2 + \cdots + x_n.$$

Here's the crux of the matter: when $p \gt 0,$ you can change the data by replacing $x_1$ by $px_1$ and $x_2$ by $x_2/p$ without changing anything you know about the data: there are still $n$ non-negative values and, because $p/p=1,$ these changes do not change the product and therefore their geometric mean remains the same.

But what happens to the sum of the numbers? Because the first value was increased by $px_1-x_1$ and the second was increased by $x_2/p - x_2,$ the change in the sum is

$$S \to S + (p-1)x_1 + (1/p-1)x_2.$$

This difference could be nonzero. But exactly what can we say about it, given we know so little about $x_1$ and $x_2$? For instance, if your favorite number is $m,$ could you make the sum equal to $nm$ (and therefore make the mean equal to $m$)? For this to happen, you need

$$nm = S + (p-1)x_1 + (1/p-1)x_2.$$

Provided $nm$ is no less than $S,$ this quadratic equation in $p$ has a solution

$$p = \frac{1}{2x_1}\left((x_1 + x_2) + (nm - S) + \sqrt{(x_1-x_2)^2 + (nm - S)^2 + 2(x_1+x_2)(nm - S)}\right).$$

Looking at this term by term, notice that since every one of $x_1,$ $x_1+x_2,$ and $nm-S$ is non-negative, and all squared terms are non-negative, indeed $p$ exists and is not negative. Therefore it gives a valid solution. Here's the main conclusion:

We can always increase the arithmetic mean of a set of two or more positive numbers by any amount we wish, without changing its geometric mean.

Finally, the AM-GM inequality states that the arithmetic mean is never less than the geometric mean and they are equal exactly when all the data are equal to their common mean. Thus, you could start with a possible dataset of all values of $10$ (your GM) and find the value of $p$ that produces a new dataset (with values of $10p, 10/p,$ and the rest remain $10$s) whose arithmetic mean equals your favorite number $m$--provided $m$ is at least $10.$

The best we can say, then, is that when $n\ge 2,$ the arithmetic mean of the data is some value $10$ or greater, but it is impossible otherwise to determine what it is.


As an example, suppose your dataset has $n=4$ elements and your favorite number larger than $10$ is $m=18.$ The formula gives $p=5,$ telling you to change the initial dataset $(10,10,10,10)$ to $(50,2,10,10).$ Sure enough, the geometric mean stays at $10$ but the arithmetic mean is $(50+2+10+10)/4=18,$ as promised.

$\endgroup$
-2
$\begingroup$

Isn't the key to answering this problem found in organizing the possible solution sets around the number of the rows in the data set (r) - no matter what that number is! - and then drawing inferences on what the data set values can be?

"For the data set the geometric mean is 10..." means if there are "r" rows in the data set, the "geometric total" for column x is 10^r. Which then yields '10' as the rth root of the geometric total, or the geometric mean.

Therefore, you're going to have values x(1), x(2), x(3),...x(r) which, when multiplied together, yield the geometric total of 10^r. That is 100, 1000, 10000, ... or some other power of 10. There is no x that, when multiplied by y <> 10, yields 10x. The only way to get “any number raised to a power of r” to equal “10 raised to a power of r” is for x to equal 10 (x^r = 10^r, therefore x=10 for all r.)

So it's not just one x value = 10, but all x values = 10 (from every row in the data set) that satisfies the computation for the geometric mean (rth root of 10^r).
But then also the sum of those x values (all x = 10), when divided by r, gives you the arithmetic mean.

The only way to get SUM(i=1...r) of x(i) = 10*r would again, be for x=10. ((x*r) / r) = r√(x^r) iff x=10.

And with { x(1) + x(2) + x(3) + ... + x(r) } also being a multiple of 10 (r*10), when divided by r we get a mean of 10 as well.

It’s just that every x = 10, and there’s no variation / deviation from the arithmetic mean.

This works also if you’re dealing with r number of values in a set, rather than r rows in a table.

Finally, if negatives are included (specifically, -10) then the arithmetic mean is unknown. Even with knowing the value of r (number of occurrences), the number of positive and negative 10’s don’t have to balance, and you could end up with an imaginary root (in the case of an odd r), and/or a case of a sum that is not evenly divisible by 10.

$\endgroup$
  • 2
    $\begingroup$ I find this argument extremely confusing because the question and its answer have nothing to do with how the numbers are represented (decimal, octal, or whatever). It's purely a matter of arithmetic. $\endgroup$ – whuber Dec 27 '18 at 20:06
  • $\begingroup$ The base representation of the numbers was only an afterthought / incidental comment. The key is that 10 is the only number that can satisfy x = 10 for all x = rth root of 10^r = sum of r x's divided by r $\endgroup$ – AndyH Dec 29 '18 at 1:25
  • $\begingroup$ I'm sorry, but I still don't see how that observation is even relevant to the question. Are you trying to show that when the AM and GM are equal, then all the numbers are equal to each other? $\endgroup$ – whuber Dec 29 '18 at 18:08
  • $\begingroup$ The base representation is not relevant. It's an incidental afterthought. I'll edit my answer. $\endgroup$ – AndyH Dec 30 '18 at 12:41
  • $\begingroup$ I think @whuber has found a general case that includes the specific case of x=r=10 that I described. $\endgroup$ – AndyH Dec 30 '18 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.