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

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

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Answer 1 · 2015-11-28T15:09:26

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

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

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

Explanation:

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

If the common ratio is $r$ $r$ then:

$a = 4 r$ $a = 4r$

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

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

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

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

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

$b = a r = 8 \sqrt[3]{2} \cdot 2 \sqrt[3]{2} = 16 \sqrt[3]{4}$ $b = ar = 8root(3)(2) * 2root(3)(2) = 16root(3)(4)$

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

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

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

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