Question

If three unequal numbers p, q, r are in HP and their squares are in AP, 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. $1+\sqrt{3}:-2:1-\sqrt{3}$

Answer 1

Answer: (A), (D)

$1-\sqrt{3}:-2:1+\sqrt{3}$
$1+\sqrt{3}:-2:1-\sqrt{3}$

By hypothesis,
$q=\frac{2pr}{p+r}$
$\Rightarrow \frac{q}{2}=\frac{pr}{p+r}=k\left(say\right)$
$\Rightarrow q=2k,pr=k(p+r)$ ....... $\left(i\right)$
Also, $p^2,q^2,r^2$ are in A.P.
$\therefore 2q^2=p^2+r^2=(p+r{)}^2-2pr$
$\Rightarrow 8k^2=(p+r{)}^2-2k(p+r)$ ...... From $\left(i\right)$
$\Rightarrow (p+r{)}^2-2(p+r)k-8k^2=0$
$\Rightarrow p+r=4k,-2k$
$\Rightarrow p+r=4k,pr=4k^2$ ...... $[\because pr=k(p+r\left)\right]$
and we get $(p-r{)}^2=(p+r{)}^2-4pr=0$
$\therefore p=r$
This is against the hypothesis
$\therefore p+r=-2k,pr=-2k^2$ ...... $[\because pr=k(p+r\left)\right]$
Now, $(p-r{)}^2=12k^2\Rightarrow p-r=\pm 2\sqrt{3}k$
Combine it with $p+r=-2k$ to get
$p=(-1\pm \sqrt{3})k$ and $r=(-1\mp \sqrt{3})k$
$\therefore p:q:r::-1\pm \sqrt{3}:2:-1\mp \sqrt{3}$
or $p:q:r::-1\mp \sqrt{3}:2:1\pm \sqrt{3}$