Question

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

Answer 1

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

`a = 8root(3)(2)`

`b = 16root(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 = 4r`

`b = ar = 4r^2`

`64 = br = 4r^3`

So `r^3 = 64/4 = 16 = 2^4`

The only Real solution to this is `r = root(3)(2^4) = 2root(3)(2)` giving:

`a = 4r = 8root(3)(2)`

`b = ar = 8root(3)(2) * 2root(3)(2) = 16root(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 root(3)(2)` and `2 omega^2 root(3)(2)`, where `omega = -1/2 + sqrt(3)/2 i` is the primitive Complex cube root of `1`.