How do you write a sequence that has three geometric means between 256 and 81?

Question

7926 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-20T23:39:32

$256, 192, 144, 108, 81$

$256=4^4$ and $81 = 3^4$

So the geometric mean of $256$ and $81$ is:

$sqrt(256*81) = sqrt(4^4*3^4) = 4^2*3^2 = 144$

The geometric mean of $256$ and $144$ is:

$sqrt(256*144) = sqrt(4^4 4^2 3^2) = 4^3*3 = 192$

The geometric mean of $144$ and $81$ is:

$sqrt(144*81) = sqrt(4^2 3^2 3^4) = 4*3^3 = 108$

The sequence: $256, 192, 144, 108, 81$ is a geometric sequence with common ratio $3/4$