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

1 Answer
Nov 28, 2015

Answer:

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#.