Explanation

(a)
Since, $p,\ q,\ r$ are positive and are in AP.
$\therefore\ \ q = \frac{p + r}{2}$ ……(i)
$\because\$ The roots of the equation $px^{2} + qx + r = 0$ are real.
$\Rightarrow \ q^{2} \geq 4pr\ \Rightarrow \ \left\lbrack \frac{p + r}{2} \right\rbrack^{2} \geq 4pr$ [from Eq. (i)]
$\Rightarrow \ p^{2} + r^{2} - 14pr\ \geq 0$
$\Rightarrow \ \left( \frac{r}{p} \right)^{2} - 14\left( \frac{r}{p} \right) + 1 \geq 0\ \ \ \ \ \ \ \ \ \ (\because p > 0)$
$\Rightarrow \ \left( \frac{r}{p} - 7 \right)^{2} - 48 \geq 0$
$\Rightarrow \ \ \left( \frac{r}{p} - 7 \right)^{2} - \left( 4\sqrt{3} \right)^{2} \geq 0$
$\Rightarrow \ \left| \frac{r}{p} - 7 \right| \geq 4\sqrt{3}$