Question

If three unequal numbers $P,Q,R$ are in H.P. and their squares are in A.P., then the ratio $P:Q:R$ is
A. $1 - \sqrt 3 :2:1 + \sqrt 3 $
B. $1:\sqrt 2 : - \sqrt 3 $
C. $1: - \sqrt 2 :\sqrt 3 $
D. $ - 1 \pm \sqrt 3 :2: - 1 \mp \sqrt 3 $

Answer 1

Hint: Given that three unequal numbers $P,Q,R$ are in H.P. It means its reciprocals are in A.P. So, $\dfrac{2}{Q} = \dfrac{1}{P} + \dfrac{1}{R}$. Let $Q = 2k$ and $PR = k(P + R)$. Also given that the squares of the numbers ${P^2},{Q^2},{R^2}$ are in A.P. So, $2{Q^2} = {P^2} + {R^2}$. Substitute $Q = 2k$ and $PR = k(P + R)$ in this equation. After substitution a quadratic formula in $\left( {P + R} \right)$ will be obtained. Solving this equation two roots will be obtained. For two different roots, two different values of $P,Q,R$ will be obtained and hence you can find the required ratio.

Complete step by step solution:
Here $P,Q,R$ are in H.P., so
$\dfrac{2}{Q} = \dfrac{1}{P} + \dfrac{1}{R}$
Add the fractions in the right hand side of the equation.
$ \Rightarrow \dfrac{2}{Q} = \dfrac{{P + R}}{{PR}}$
Take reciprocal of both the expressions on both sides of the equation.
$ \Rightarrow \dfrac{Q}{2} = \dfrac{{PR}}{{P + R}}$
Let $\dfrac{Q}{2} = \dfrac{{PR}}{{P + R}} = k$(say)
Then $\dfrac{Q}{2} = k$ and $\dfrac{{PR}}{{P + R}} = k$
$Q = 2k$ and $PR = k(P + R).....(i)$
Again, the squares of the numbers i.e. ${P^2},{Q^2},{R^2}$ are in A.P. So,
$2{Q^2} = {P^2} + {R^2}$
Use the formula ${a^2} + {b^2} = {(a + b)^2} - 2ab$
$ \Rightarrow 2{Q^2} = {\left( {P + R} \right)^2} - 2PR$
Substitute $Q = 2k$ and $PR = k\left( {P + R} \right)$
$ \Rightarrow 8{k^2} = {\left( {P + R} \right)^2} - 2k\left( {P + R} \right)$
This is a quadratic equation in $\left( {P + R} \right)$
Let $P + R = S$
Then the equation becomes $8{k^2} = {S^2} - 2kS$
Arrange the terms of the equation
$ \Rightarrow {S^2} - 2kS - 8{k^2} = 0.....(ii)$
Factorize the expression on the left hand side of the equation.
$\begin{array}{l}{S^2} - 2kS - 8{k^2}\\ = {S^2} - 4kS + 2kS - 8{k^2}\\ = S\left( {S - 4k} \right) + 2k\left( {S - 4k} \right)\\ = \left( {S - 4k} \right)\left( {S + 2k} \right)\end{array}$
From equation $(ii)$, we get
$ \Rightarrow \left( {S - 4k} \right)\left( {S + 2k} \right) = 0$
If $S - 4k = 0$, then $S = 4k$
If $S + 2k = 0$, then $S = - 2k$
We assumed that $P + R = S$
So, $P + R = 4k$ or $P + R = - 2k$
Substitute the expressions for $\left( {P + R} \right)$ in equation $(i)$
If $P + R = 4k$, then $PR = 4{k^2}$
If $P + R = - 2k$, then $PR = - 2{k^2}$
Now, use the formula ${\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab$
$\therefore {\left( {P - R} \right)^2} = {\left( {P + R} \right)^2} - 4PR$
If $P + R = 4k$, then $PR = 4{k^2}$, so
$\begin{array}{l}{\left( {P - R} \right)^2} = 16{k^2} - 16{k^2}\\ \Rightarrow {\left( {P - R} \right)^2} = 0\\ \Rightarrow P - R = 0\\ \Rightarrow P = R\end{array}$
But it is given that $P,Q,R$ are not equal. So, it is impossible.
Now, we have only one option i.e. $P + R = - 2k...(iii)$, then $PR = - 2{k^2}$
Using these, we get
$\begin{array}{l}{\left( {P - R} \right)^2} = 4{k^2} + 8{k^2}\\ \Rightarrow {\left( {P - R} \right)^2} = 12{k^2}\\ \Rightarrow P - R = \pm \sqrt {12{k^2}} \\ \Rightarrow P - R = \pm 2\sqrt 3 k.....(iv)\end{array}$
Taking $P + R = - 2k$ and $P - R = 2\sqrt 3 k$, we get
$\begin{array}{l}2P = - 2k + 2\sqrt 3 k\\ \Rightarrow P = - k + \sqrt 3 k\\ \Rightarrow P = \left( { - 1 + \sqrt 3 } \right)k\end{array}$
Putting the value of $P$ in equation $(iii)$, we get
$\begin{array}{l}\left( { - 1 + \sqrt 3 } \right)k + R = - 2k\\ \Rightarrow R = - 2k - \left( { - 1 + \sqrt 3 } \right)k\\ \Rightarrow R = \left( { - 2 + 1 - \sqrt 3 } \right)k\\ \Rightarrow R = \left( { - 1 - \sqrt 3 } \right)k\end{array}$
∴ We have $P = \left( { - 1 + \sqrt 3 } \right)k$ and $R = \left( { - 1 - \sqrt 3 } \right)k$ and $Q = 2k$
Hence the ratio $P:Q:R = \left( { - 1 + \sqrt 3 } \right)k:2k:\left( { - 1 - \sqrt 3 } \right)k = \left( { - 1 + \sqrt 3 } \right):2:\left( { - 1 - \sqrt 3 } \right).....(v)$
Taking $P + R = - 2k$ and $P - R = - 2\sqrt 3 k$, we get
$\begin{array}{l}2P = - 2k - 2\sqrt 3 k\\ \Rightarrow P = - k - \sqrt 3 k\\ \Rightarrow P = \left( { - 1 - \sqrt 3 } \right)k\end{array}$
Putting the value of P in equation iii, we get
$\begin{array}{l}\left( { - 1 - \sqrt 3 } \right)k + R = - 2k\\ \Rightarrow R = - 2k - \left( { - 1 - \sqrt 3 } \right)k\\ \Rightarrow R = \left( { - 2 + 1 + \sqrt 3 } \right)k\\ \Rightarrow R = \left( { - 1 + \sqrt 3 } \right)k\end{array}$
We have $P = \left( { - 1 - \sqrt 3 } \right)k$ and $R = \left( { - 1 + \sqrt 3 } \right)k$ and $Q = 2k$
Hence the ratio $P:Q:R = \left( { - 1 - \sqrt 3 } \right)k:2k:\left( { - 1 + \sqrt 3 } \right)k = \left( { - 1 - \sqrt 3 } \right):2:\left( { - 1 + \sqrt 3 } \right).....(vi)$
From the ratios obtained in $(v)$ and $(vi)$, we get
The ratio $P:Q:R$ is $\left( { - 1 \pm \sqrt 3 } \right):2:\left( { - 1 \mp \sqrt 3 } \right)$

Note: Some terms are in H.P. means the reciprocals of the terms are in A.P.
It is given in the question that the numbers $P,Q,R$ are unequal. For this reason, $P = R$ has been rejected.
The obtained value of $P$ is $\left( { - 1 + \sqrt 3 } \right)$ for $R = - 1 - \sqrt 3 $ and the obtained value of $P$ is $\left( { - 1 - \sqrt 3 } \right)$ for $R = - 1 + \sqrt 3 $. You should keep it in mind while writing the answer.