Question

Three unequal numbers are in H.P. and their squares are in A.P. prove that numbers are in the ratio
$1-\sqrt{3}:-2:1+\sqrt{3}or1+\sqrt{3}:-2:1-\sqrt{3}$

Answer 1

Let the numbers be x, y and z
$\therefore y=\frac{2xz}{x+z}ory(x+z)=2xz$ ...(1)
Again$x^2,y^2,z^2$ are in A.P.
$2y^2=x^2+z^2=(x+z{)}^2-2xz$
$2y^2=(x+z{)}^2-y(x+z),by(1)$or
or $(x+z{)}^2-y(x+z)-2y^2=0$
$\therefore x+z=\frac{y\pm \sqrt{y^2+8y^2}}{2}$
or x +z = 2y or -y
Case(I) x + z = 2y .....(2)
Eliminate y between (1) and (2)
$(x+z{)}^2=4xz$
or $(x-z{)}^2=0\therefore x=z$
Above is not true as the numbers are distinct.
Case (II) x + z = -y ...(3)
Eliminate y between (1) and (3)
$-(x+z{)}^2=2xz$
$\therefore ,x^2+4xz+z^2=0or{\left(\frac{x}{z}\right)}^2+4\frac{x}{z}+1=0$
$\therefore \frac{x}{z}=-2\pm \sqrt{3}$
Choose $\frac{x}{z}=-2+\sqrt{3}$
$\therefore From\left(3\right),\frac{-y}{2}=1+\frac{x}{z}=-1+\sqrt{3}$
or $\frac{y}{z}=1-\sqrt{3}$
$\frac{x}{\sqrt{3}-2}=\frac{y}{1-\sqrt{3}}=\frac{z}{1}$ ....(4)
Similarly if $\frac{x}{z}=-2-\sqrt{3}then\frac{y}{z}=1+\sqrt{3}$
$\therefore \frac{x}{-(\sqrt{3}-2)}=\frac{y}{1-\sqrt{3}}=\frac{z}{1}$
The given ratio $1-\sqrt{3}:-2:1+\sqrt{3}$ is same
or $\frac{1-\sqrt{3}}{1+\sqrt{3}}:\frac{-2}{1+\sqrt{3}}:1=\frac{4-2\sqrt{3}}{1-3}:\frac{-2(1-\sqrt{3})}{1-3}:1$
$=\sqrt{3}-2:1-\sqrt{3}:1$ same as (4) etc.
Again taking $\frac{x}{z}=-2-\sqrt{3},$ we will get the second answer