Find the ratio $p:q:r$, if $p,q,r$ are in H.P, and their squares are in A.P.

Question

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?

Answer 1

$ q= \frac{2pr}{p+r} $

$ 2pr=q(p+r) $.......(1)

$ p^2, q^2, r^2 $ are in AP thus

$ (p+r)^2 -2pr=2q^2 $

$ (p+r)^2 -(p+r)q=2q^2 $.......(2)

Solve using quadratic equation.

You will get two cases, one will be $ p + r = -q $ and other will be $ p + r = 2q $

Solve each case with equation (1) and (2).

Reject $ p + r = 2q $.

Solving $ p + r = -q $ and (1) and (2) you will get $ p - r = \pm \sqrt3 q $. Now solving this with $ p + r = -q $ you will get relation between $ p,q,r $. Solving this you will get the ratio.

Will you take it from here?? Or should I solve it further?

Answer 2

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 $x = -2-\sqrt{3}$.