0
$\begingroup$

Is there, at all any relationship between the geometric mean of a set and the arithmetic mean of a set? If I knew one, and I knew the number of terms in the set, how could I calculate the other?

$\endgroup$
2
0
$\begingroup$

In general, there is no relationship between the arithmetic mean and the geometric mean of the same data set other than the fact that AM is always greater than or equal to GM, with equality if and only if all of the numbers in the data set are equal.

In fact, given any two positive numbers $a$ and $g$ with $a \ge g$, there will always be two numbers whose arithmetic and geometric means are $a$ and $g$ respectively. Indeed, to find those two numbers, it suffices to solve the equation $x(2a-x)=g^2$. This equation can be rearranged as $-x^2+2ax-g^2=0$. Using the quadratic formula, the two solutions for $x$ are $-\frac{1}{2}(-2a \pm \sqrt{4a^2-4g^2})$. Those two solutions will then have AM $a$ and GM $g$.

Also, it is true that the logarithm of a geometric mean is the arithmetic mean of the logarithms. Conversely, applying an exponential function to an arithmetic mean gives the geometric mean of the values of the exponential function applied to all the members of the data set.

$\endgroup$
2
  • $\begingroup$ Why solve x(2a-x)=g^2? $\endgroup$
    – 4yl1n
    Feb 8 '20 at 22:06
  • $\begingroup$ Because if $(x+y)/2=a$ and $\sqrt{xy}=g$, then $y=2a-x$ and $x(2a-x)=xy=g^2$. $\endgroup$ Feb 8 '20 at 23:13

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.