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There are some geometric means between $\dfrac {1}{2}$ and $16$. If the third mean be $4$, find the number of means between two numbers. Also find the last mean.

My Attempt Let $N$ be the number of geometric means between $\dfrac {1}{2}$ and $16$. $$a=\dfrac {1}{2}$$ $$b=16$$ Now, $$r=(\dfrac {b}{a})^{\dfrac {1}{N+1}}$$ $$r=2^{\dfrac {5}{N+1}}$$

Then, What's next!?

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    `the third mean be 4` You need to use this somewhere along the way. – 2017-02-13
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    @ dxiv, using that I got $\textrm N=\dfrac {2}{3}$$.. I think it's wrong – 2017-02-13
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    Doublecheck your calculations. That condition should give the common ratio as $\sqrt[3]{\left(4 / \frac{1}{2}\right)}\,$. – 2017-02-13
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    @ dxiv, then $N=4$. – 2017-02-13
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    If the common ratio is $2$ and the $3^{rd}$ term is $4$ then $16$ will be the $5^{th}$ term. If what the problem calls the `last mean` is the term just before that then, yes, it would be the $4^{th}\,$. – 2017-02-13
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    @ dxiv, Last mean is the term preceding the last term – 2017-02-13
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    Then you got it right, that's the $4^{th}$. – 2017-02-13

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The means will be $\dfrac 12r, \dfrac 12r^2, \dfrac 12r^3, \dots, \dfrac 12r^n$ for some $n$. If the third mean is $4$, then

\begin{align} \dfrac 12r^3 &= 4\\ r^3 &= 8 \\ r &= 2 \end{align}

So the means are $1,2,4,8$