Explanation

(d) Let $R$ be the equivalent resistance. Then addition/ subtraction of one more seat of resistors $R_{1},\ R_{2}$and $R_{3}$will not affect the total resistance. Thus,
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$R = R_{1} +$(parallel combination of $R$ and $R_{2}$)$+ R_{3}$
$R = R_{1} + \left( \frac{RR_{3}}{R + R_{3}} \right) + R_{2}$
$\Rightarrow \ R^{2} + RR_{3} = RR_{1} + R_{1}R_{3} + RR_{3} + RR_{2} + R_{2}R_{3}$
$\Rightarrow \ R^{2} - R\left( R_{1} + R_{2} \right) = \left( R_{2} + R_{1} \right)R_{3} = 0$
$\Rightarrow \ R = \frac{(R_{1} + R_{2}) \pm \sqrt{\left( R_{1} + R_{2} \right)^{2} + 4\left( R_{1} + R_{2} \right)R_{3}}}{2}$
As $R$ cannot be negative, hence
$R = \frac{1}{2}\left\lbrack (R_{1} + R_{2} + \sqrt{\left( R_{1} + R_{2} \right)^{2} + 4\left( R_{1} + R_{2} \right)R_{3}} \right\rbrack$
$= \frac{1}{2}\left\lbrack \left( R_{1} + R_{2} \right) + \sqrt{\left( R_{1} + R_{2} \right)(R_{1} + R_{2} + 4R_{3})} \right\rbrack$