Explanation

(d)
Since, $p,q\ and\ r$ in HP.
$\Rightarrow q = \frac{2pr}{p + r}\ \Rightarrow \frac{q}{2} = \ \frac{pr}{p + r} = K$ (say)
$\Rightarrow q = 2K,\ pr = (p + r)K$
Also, $p^{2},\ q^{2},\ r^{2}$ are in AP.
$\therefore\ 2q^{2} = p^{2} + r^{2} = (p + r)^{2} - 2pr$
$\Rightarrow 8K^{2} = (p + r)^{2} - 2(p + r)K$
$\Rightarrow (p + r)^{2} - 2(p + r)K - 8K^{2} = 0$
$\Rightarrow p + r = 4K,\ - 2K$
When $p + r = 4K,$ then $pr = 4K^{2}$
$\therefore(p - r)^{2} = (p + r)^{2} - 4pr = 16K^{2} - 16K^{2} = 0$
$\Rightarrow \ p = r$
But this is not possible $(\because p \neq r)$
$\therefore p + = - 2K\ \Rightarrow pr = - 2K \bullet K = - 2K^{2}$
Now, $(p - r)^{2} = (p + r)^{2} - 4pr$
$= 4K^{2} - 4\left( - 2K^{2} \right) = 12\ K^{2}$
$\Rightarrow p - r = \pm 2\sqrt{3}K$
$\Rightarrow p = \left( - 1 \pm \sqrt{3} \right)K$
And $2r = - 2K \mp \sqrt{3}K$
$\Rightarrow \ r = \left( - 1 \mp \sqrt{3} \right)K$
$\therefore p\ :q\ :r = \left( - 1 \mp \sqrt{3} \right)K\ :2K\ :\left( - 1 \mp \sqrt{3} \right)K$
$= - 1 \mp \sqrt{3}\ :2: - \ 1 \mp \sqrt{3}$
$= \$($- 1 \mp \sqrt{3}$ : ($-$2) : ($- 1 \mp \sqrt{3}$)