# How do you find two geometric means between 4 and 64?

Nov 28, 2015

Find a geometric sequence $4$, $a$, $b$, $64$ by solving for the common ratio to find:

$a = 8 \sqrt[3]{2}$

$b = 16 \sqrt[3]{4}$

#### Explanation:

We are effectively being asked to find $a$ and $b$ such that $4$, $a$, $b$, $64$ is a geometric sequence.

If the common ratio is $r$ then:

$a = 4 r$

$b = a r = 4 {r}^{2}$

$64 = b r = 4 {r}^{3}$

So ${r}^{3} = \frac{64}{4} = 16 = {2}^{4}$

The only Real solution to this is $r = \sqrt[3]{{2}^{4}} = 2 \sqrt[3]{2}$ giving:

$a = 4 r = 8 \sqrt[3]{2}$

$b = a r = 8 \sqrt[3]{2} \cdot 2 \sqrt[3]{2} = 16 \sqrt[3]{4}$

Then $a$ will be the geometric mean of $4$ and $b$, and $b$ will be the geometric mean of $a$ and $64$

The other possible common ratios that work are $2 \omega \sqrt[3]{2}$ and $2 {\omega}^{2} \sqrt[3]{2}$, where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.