A number $a$ is the Arithmetic Mean(A.M.) between $b$ and $c$, $b$ is the Geometric Mean(G.M.) between $a$ and $c$. Prove that $$\frac{1}{a}, \frac{1}{c} and \frac{1}{b}$$ are in Arithmetic Progressions(A.P.).
I haven't been able to do much, but this is it :
As, $a$ is the A.M. between $b$ and $c$, $$a = \frac{b+c}{2}$$ or, $$2a = b+c$$
And, $b$ is the G.M. between $a$ and $c$, so, $$b^2 = ac$$ or, $$b = \sqrt{ac}$$How do i proceed next?
As $a$ is the Arithmetic mean, it lies between $b$ and $c$. But, as $b$ is the geometric mean it will lie between $a$ and $c$. This can only mean that $a=b$. And because $a=\frac{b+c}{2}$, $a=b=c$. Clearly,$\frac{1}{a}= \frac{1}{c} = \frac{1}{b}$ and therefore, trivially, in AP.
Write $b = 2a - c$ so that $b^2 = (2a - c)^2$. Then with the additional condition $b^2 = ac$, eliminate $b$ and obtain an equation in $a, c$. Factor this equation and consider two cases, one of which is $a = c$.
