Explanation

(d) Let $R_{0}$ be the initial resistance of both conductors.
$\therefore$ At temperature $\theta$ their resistances will be,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R_{1} = R_{0}(1 + \alpha_{1}\theta$)
and $\ \ \ \ \ \ \ \ \ \ R_{2} = R_{0}(1 + \alpha_{2}\theta)$
For series combination, $R_{s} = R_{1} + R_{2}$
$R_{s}\left( 1 + \alpha_{s}\theta \right) = R_{0}\left( 1 + \alpha_{1}\theta \right) + R_{0}(1 + \alpha_{2}\theta)$
Where, $\ \ \ R_{s0} = R_{0} + R_{0} = 2R_{0}$
$\therefore 2R_{0}\left( 1 + \alpha_{2}\theta \right) = 2R_{0} + R_{0}\theta(\alpha_{1} + \alpha_{2})$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha_{s} = \frac{\alpha_{1} + \alpha_{2}}{2}$
For parallel combination,
$R_{p} = \frac{R_{1}R_{2}}{R_{1} + R_{2}}$
$R_{p0}\left( 1 + \alpha_{p}\theta \right) = \frac{R_{0}\left( 1 + \alpha_{1}\theta \right)R_{0}(1 + \alpha_{2}\theta)}{R_{0}\left( 1 + \alpha_{1}\theta \right) + R_{0}(1 + \alpha_{2}\theta)}$
Where, $R_{p0} = \frac{R_{0}R_{0}}{R_{0} + R_{0}} = \frac{R_{0}}{2}$
$\therefore\ \ \frac{R_{0}}{2}\left( 1 + \alpha_{p}\theta \right) = \frac{R_{0}^{2}(1 + \alpha_{1}\theta + \alpha_{2}\theta + \alpha_{1}\alpha_{2}\theta^{2})}{R_{0}(2 + \alpha_{1}\theta + \alpha_{2}\theta)}$
as $\alpha_{1}\ and\ \alpha_{2}$ are small quantities.
$\therefore\ \alpha_{1}\alpha_{2}$ is negligible.
or $\alpha_{p} = \frac{\alpha_{1} + \alpha_{2}}{2 + (\alpha_{1} + \alpha_{2})\theta} = \frac{\alpha_{1} + \alpha_{2}}{2}\left\lbrack 1 - \left( \frac{\alpha_{1} + \alpha_{2}}{2} \right)\theta \right\rbrack$
as $\left( \alpha_{1} + \alpha_{2} \right)^{2}\ \ $is negligible
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \alpha_{p} = \frac{\alpha_{1} + \alpha_{2}}{2}$