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Three unequal numbers are in Harmonic Progression and their squares are in Arithmetic progression. Prove that the numbers are in the ratio 1-3^1/2:-2:1+3^1/2

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You've already tried doing anything for this problem? That should be included in your question. –  J. M. Jul 23 '12 at 6:51

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Three numbers are in harmonic progression if their reciprocals are in arithmetic progression. Call the numbers $x$, $y$, and $z$, with the "middle" one being $y$. Our harmonic progression requirement can be written as $$\frac{1}{y}-\frac{1}{x}=\frac{1}{z}-\frac{1}{y}$$ (the difference of successive reciprocals is constant).

Because the squares are in arithmetic progression, we have $$y^2-x^2=z^2-y^2.$$ We can choose one of the numbers almost arbitrarily. So let $y=1$. (If you don't like this, you can keep $y$. If we want to get the answer as given directly, let $y=-2$.) With a little simplification, our two equations can be written as $$2xz=x+z \qquad\text{and}\qquad x^2+z^2=2.$$ Now we need to solve for $x$ and $z$. There are various approaches. We use one that is useful elsewhere.

Let $p=xz$ and $s=x+z$. Here $p$ of course stands for product, and $s$ stands for sum.

The first equation is simply $2p=s$. For the second equation, use $x^2+z^2=(x+z)^2-2xz=s^2-2p$. So the second equation is $s^2-2p=2$. But $2p=s$. Substituting, we get $s^2-s-2=0$. This is easy to solve for $s$. We get $s=-1$ or $s=2$.

Suppose first that $s=-1$. Then $p=-\frac{1}{2}$. So $x$ and $z$ satisfy the quadratic equation $$w^2+w-\frac{1}{2}.$$ Solve. We get $$w=\frac{-1\pm\sqrt{3}}{2}.$$ This gives the ratios $$\frac{-1-\sqrt{3}}{2}:1:\frac{-1+\sqrt{3}}{2}.$$ Multiply by $-2$ and reverse the order to get the answer asked for.

Finally, look at the possibility $s=2$. That gives $p=1$, and fairly quickly $x=z$, contradicting the requirement of distinctness.

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