# How do you find 3 geometric means between 3 and 1488 ?

Jul 25, 2016

$3 , \textcolor{red}{6 \sqrt{31}} , \textcolor{g r e e n}{12 \sqrt{31}} , \textcolor{b l u e}{24 {\left(\sqrt{31}\right)}^{3}} , 1488$

#### Explanation:

We are looking for $3$ numbers $\textcolor{red}{a} , \textcolor{g r e e n}{b} , \textcolor{b l u e}{c}$ such that:

$\textcolor{red}{a}$ is the geometric mean of $3$ and $\textcolor{g r e e n}{b}$

$\textcolor{g r e e n}{b}$ is the geometric mean of $\textcolor{red}{a}$ and $\textcolor{b l u e}{c}$

$\textcolor{b l u e}{c}$ is the geometric mean of $\textcolor{g r e e n}{b}$ and $1488$

That will make the following sequence into a geometric one:

$3 , \textcolor{red}{a} , \textcolor{g r e e n}{b} , \textcolor{b l u e}{c} , 1488$

If the common ratio is $r$ then we must have:

$1488 = 3 {r}^{4}$

So:

${r}^{4} = \frac{1488}{3} = 496 = {2}^{4} \cdot 31$

in order that the geometric means be Real and positive, we need to choose the principal $4$th root to find:

$r = 2 \sqrt{31}$

Hence $\textcolor{red}{a} , \textcolor{g r e e n}{b} , \textcolor{b l u e}{c}$ are:

$3 \cdot 2 \sqrt{31} = \textcolor{red}{6 \sqrt{31}}$

$6 \sqrt{31} \cdot 2 \sqrt{31} = \textcolor{g r e e n}{12 \sqrt{31}}$

$12 \sqrt{31} \cdot 2 \sqrt{31} = \textcolor{b l u e}{24 {\left(\sqrt{31}\right)}^{3}}$