<s>[INST] Find the ratio $p:q:r$, if $p,q,r$ are in H.P, and their squares are in A.P. 
If three unequal numbers $p,q,r$ are in H.P and their squares are in A.P, then find the ratio $p:q:r$
  .

Attempt
A.P(1): $\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$
$$
\dfrac{1}{q}-\dfrac{1}{p}=\dfrac{1}{r}-\dfrac{1}{q}\implies\dfrac{p+r}{pr}=\dfrac{2}{q}\\
$$
A.P(2): $p^2,q^2,r^2$
$$ q^2-p^2=r^2-q^2\implies p^2+r^2=2q^2\\
$$
As this was asked as a multiple choice question i was wondering what is the easiest way to solve this?
 [/INST] Start with your
$\dfrac{1}{q}-\dfrac{1}{p}=\dfrac{1}{r}-\dfrac{1}{q}\implies\dfrac{p+r}{pr}=\dfrac{2}{q}\\
q^2-p^2=r^2-q^2\implies p^2+r^2=2q^2\\
$
Then
$q= \dfrac{2pr}{p+r}$.
Putting this to the second equation,
$p^2+r^2
= 2(\dfrac{2pr}{p+r})^2
=\dfrac{8p^2r^2}{(p+r)^2}
$
or
$(p^2+r^2)(p+r)^2
=8p^2r^2
$.
Dividing by $r^4$
and letting
$x = p/r$,
$(1+x^2)(1+x)^2
=8x^2$.
With the help of Wolfy,
this is
$0 = (1-x)^2(x^2+4x+1)
$
so
$x = 1$
or
$x
=\dfrac{-4\pm\sqrt{12}}{2}
=-2\pm \sqrt{3}
$.
If $x=1$ then
$p = r = q$.
If $x = -2+\sqrt{3}$,
$p = r(-2+\sqrt{3})$
so
$\begin{array}\\
q
&=\dfrac{2r^2(-2+\sqrt{3})}{r+r(-2+\sqrt{3})}\\
&=r\dfrac{2(-2+\sqrt{3})}{-1+\sqrt{3}}\\
&=r(1+\sqrt{3})(-2+\sqrt{3})
\qquad\text{since }(-1+\sqrt{3})(1+\sqrt{3})=2\\
&=r(1-\sqrt{3})\\
\end{array}
$
If $x = -2-\sqrt{3}$,
$p = r(-2-\sqrt{3})$
so
$\begin{array}\\
q
&=\dfrac{2r^2(-2-\sqrt{3})}{r+r(-2-\sqrt{3})}\\
&=r\dfrac{2(-2-\sqrt{3})}{-1-\sqrt{3}}\\
&=-r\dfrac{2(-2-\sqrt{3})}{1+\sqrt{3}}\\
&=-r(-1+\sqrt{3})(-2-\sqrt{3})
\qquad\text{since }(-1+\sqrt{3})(1+\sqrt{3})=2\\
&=r(-1+\sqrt{3})(2+\sqrt{3})\\
&=r(1+\sqrt{3})\\
\end{array}
$
Check (with $r = 1$)
with $x = -2+\sqrt{3}$.
$p+r = -1+\sqrt{3}
$
and
$pr =-2+\sqrt{3}
$
so
$2pr  =-4+2\sqrt{3}
$
and
$q(p+r)
=(1-\sqrt{3})(-1+\sqrt{3})
=-4+2\sqrt{3}
$
and
$q^2 = 4-2\sqrt{3}$
and
$p^2+r^2
=1+(-2+\sqrt{3})^2
=1+7-4\sqrt{3}
=8-4\sqrt{3}
$.
It also checks for
 $x = -2-\sqrt{3}$.
</s>