Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.