# How do you prove that arithmetic mean, geometric mean and harmonic mean of two numbers are in geometric sequence?

Jan 31, 2017

See explanation.

#### Explanation:

Let's let the two numbers in question be $a$ and $b$.

The arithmetic mean ${M}_{A}$ of $n$ numbers is the familiar idea of a mean—it's the sum of the numbers, divided by how many there are. In this case,

${M}_{A} = \frac{a + b}{2}$.

Since ${M}_{A}$ is halfway between $a$ and $b$, ${M}_{A}$ is just as far from $a$ as it is from $b$. In math terms:

${M}_{A} - a \text{ "=" } b - {M}_{A}$

The geometric mean ${M}_{G}$ of $n$ numbers is the product of the numbers, taken to the ${n}^{\text{th}}$ root. In this case,

${M}_{G} = \sqrt{a \times b}$

Under this definition, the proportional change from $a$ to ${M}_{G}$ is the same as the proportional change from ${M}_{G}$ to $b$. In math terms:

${M}_{G} / a = \frac{b}{M} _ G$

The harmonic mean ${M}_{H}$ of $n$ numbers is... the reciprocal of the mean of the reciprocals of the numbers. Wow, that's a mouthful. Basically, you take the reciprocals of the numbers, find their (arithmetic) mean, and take the reciprocal of this mean. In this case,

${M}_{H} = {\left[\frac{\frac{1}{a} + \frac{1}{b}}{2}\right]}^{\text{-1"" "=" }} \frac{2}{\frac{1}{a} + \frac{1}{b}}$

$\frac{1}{M} _ H$ is halfway between $\frac{1}{a}$ and $\frac{1}{b}$. In math terms:

$\frac{1}{M} _ H - \frac{1}{a} \text{ "=" } \frac{1}{b} - \frac{1}{M} _ H$.

## On with the proof!

To prove that ${M}_{A} , {M}_{G} ,$ and ${M}_{H}$ are in geometric sequence, we need to show that there is a common ratio $R$ between them.

First, let's try ${M}_{G} / {M}_{A}$. This is equal to

${M}_{G} / {M}_{A} = \frac{\sqrt{a \times b}}{\left(a + b\right) / 2} = \frac{2 \sqrt{a b}}{a + b} \text{ "-=" } R$

(Let's denote this ratio $R$.)

Now let's try ${M}_{H} / {M}_{G}$:

${M}_{H} / {M}_{G} = \frac{2 / \left(\frac{1}{a} + \frac{1}{b}\right)}{\sqrt{a \times b}}$

$\textcolor{w h i t e}{{M}_{H} / {M}_{G}} = \frac{2}{\left(\frac{1}{a} + \frac{1}{b}\right) \sqrt{a b}} \textcolor{b l u e}{\times \frac{\sqrt{a b}}{\sqrt{a b}}}$

color(white)(M_H/M_G) = (2sqrt(ab))/((1/a+1/b)ab

$\textcolor{w h i t e}{{M}_{H} / {M}_{G}} = \frac{2 \sqrt{a b}}{\frac{a b}{a} + \frac{a b}{b}}$

$\textcolor{w h i t e}{{M}_{H} / {M}_{G}} = \frac{2 \sqrt{a b}}{b + a} \text{ "=" "(2sqrt(ab))/(a+b)" "=" } R$

And hey, presto—we're done! We've shown that ${M}_{G} / {M}_{A} = {M}_{H} / {M}_{G} = R$. This means ${M}_{A} \times R = {M}_{G}$, and ${M}_{G} \times R = {M}_{H}$, and so these 3 means are consecutive terms in a geometric sequence, in the order ${M}_{A} , {M}_{G} , {M}_{H}$, with common ratio $R = \frac{2 \sqrt{a b}}{a + b} .$