Question

If three unequal numbers $\mathrm{p},\mathrm{q},\mathrm{r}$ are in H.P. and their squares are in A.P., then the ratio $\mathrm{p}:\mathrm{q}:\mathrm{r}$ 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

The correct option is D

$-1\pm \sqrt{3}:2:-1\mp \sqrt{3}$

Explanation for the correct answer:

Step 1: Applying the rule of H.P.

As, $\mathrm{p},\mathrm{q},\mathrm{r}$ are in H.P, so $\frac{1}{p},\frac{1}{q},\frac{1}{r}$ are in A.P.

$\begin{array}{rcl}& \Rightarrow & \frac{1}{q}-\frac{1}{p}=\frac{1}{r}-\frac{1}{q}\\ \Rightarrow \frac{2}{q}& =& \frac{1}{r}+\frac{1}{p}\\ \Rightarrow \frac{2}{q}& =& \frac{p+r}{pr}.......\left(1\right)\end{array}$

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

Now,

$\begin{array}{rcl}\text{Let}\frac{pr}{p+r}& =& k\\ \Rightarrow pr& =& k\left(p+r\right)......\left(2\right)\end{array}$

and

$\begin{array}{rcl}\frac{q}{2}& =& k\\ & \Rightarrow & q=2k.....\left(3\right)\end{array}$

Step 2: Applying the rule of A.P.

As, the squares of $\mathrm{p},\mathrm{q},\mathrm{r}$ are in A.P, so

$\begin{array}{rcl}{q}^2-p^2& =& r^2-q^2\\ \Rightarrow 2q^2& =& r^2+p^2\\ \Rightarrow 2q^2& =& {\left(p+r\right)}^2-2pr\left[bytheidentity{\left(a+b\right)}^2\right]\\ \Rightarrow 2{\left(2k\right)}^2& =& {\left(p+r\right)}^2-2k\left(p+r\right)\mathbf{}\left[\mathrm{from}\left(2\right)\&\left(3\right)\right]\\ \Rightarrow 8k^2& =& {\left(p+r\right)}^2-2k\left(p+r\right)\\ & \Rightarrow & {\left(p+r\right)}^2-2k\left(p+r\right)-8k^2=0\end{array}$

Step 3: Solving the equation.

Let $p+r=a$, so the equation can be written as

$\begin{array}{rcl}{a}^2-2ak-8k^2& =& 0\\ \Rightarrow a^2+2ak-4ak-8k^2& =& 0\\ & \Rightarrow & a\left(a+2k\right)-4k\left(a+2k\right)=0\\ \Rightarrow \left(a-4k\right)\left(a+2k\right)& =& 0\\ \Rightarrow a& =& 4kor-2k\end{array}$

Step 4: Finding the values of $\mathrm{p}\&\mathrm{r}$.

Taking $a=-2k$, we get

$p+r=-2k.....\left(4\right)$

Now, we can say that

$\begin{array}{rcl}{\left(p-r\right)}^2& =& {\left(p+r\right)}^2-4pr\\ & =& {\left(-2k\right)}^2-4k\left(p+r\right)\\ & =& 4k^2-4k\left(-2k\right)\\ & =& 4k^2+8k^2\\ & =& 12k^2\end{array}$

From here,

$\begin{array}{rcl}p-r& =& \sqrt{12k^2}\\ & =& \pm 2\sqrt{3}k.....\left(5\right)\end{array}$

From $\left(4\right)+\left(5\right)$, we get

$\begin{array}{rcl}2p& =& -2k\pm 2\sqrt{3}k\\ & \Rightarrow & p=-k\pm \sqrt{3}k\\ & =& k\left(-1\pm \sqrt{3}\right)\end{array}$

By substituting the value of $p$ in $\left(4\right)$, we get

$\begin{array}{rcl}r& =& -2k-k\left(-1\pm \sqrt{3}\right)\\ & =& -2k+k\mp \sqrt{3}k\\ & =& -k\mp \sqrt{3}k\\ & =& -k(1\pm \sqrt{3})\end{array}$

Step 5: Finding the ratio of $\mathrm{p},\mathrm{q},\mathrm{r}$.

We have

$$\begin{gathered} p=k\left(-1\pm \sqrt{3}\right) \\ \begin{array}{rcl}& & q\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & k\end{array} \\ \begin{array}{rcl}& & r\end{array}\begin{array}{rcl}& =& -k(1\pm \sqrt{3})\end{array} \end{gathered}$$

$So,p:q:r=-1\pm \sqrt{3}:2:-1\mp \sqrt{3}$

Hence, the correct answer is option D.