Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are $\mathrm{\alpha }$₁ & $\mathrm{\alpha }$₂. The respective temperature coefficients of their series and parallel combinations are nearly

$T\mathrm{heir} \mathrm{intial} \mathrm{resistances} \mathrm{are} \mathrm{same}$

$$\begin{gathered} \mathrm{so} \mathrm{at} \mathrm{any} \mathrm{temperature} \mathrm{their} \mathrm{series} \mathrm{equivalent} \mathrm{will} \mathrm{be} \\ {\mathrm{R}}_{\mathrm{s}}=\mathrm{R}\left(1+{\mathrm{\alpha }}_1\mathrm{t}\right)+\mathrm{R}\left(1+{\mathrm{\alpha }}_2\mathrm{t}\right) \\ \mathrm{effective} \mathrm{temperature} \mathrm{coefficient} \\ {\mathrm{\alpha }}_{\mathrm{s}}=\frac{{\mathrm{R\alpha }}_1\mathrm{t}+{\mathrm{R\alpha }}_1\mathrm{t}}{2\mathrm{Rt}}=\frac{{\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2}{2} \end{gathered}$$

Similarly for parallel combination

$\frac{1}{R_p\left(1+{\alpha }_{p}T\right)}=\frac{1}{R\left(1+{\alpha }_{1}T\right)}+\frac{1}{R\left(1+{\alpha }_{2}T\right)}$

$\frac{2}{R\left(1+{\alpha }_{p}T\right)}=\frac{2+\left({\alpha }_1+{\alpha }_2\right)T}{R\left(1+\left({\alpha }_1+{\alpha }_2\right)T+{\alpha }_1{\alpha }_2{T}^2\right)}$

$1+{\alpha }_{\mathrm{p}}\mathrm{T}=1+\frac{\left({\alpha }_1+{\alpha }_2+2{\alpha }_1{\alpha }_2\mathrm{T}\right)\mathrm{T}}{2+\left({\alpha }_1+{\alpha }_2\right)\mathrm{T}}$

Since temperature coefficient of resistance is small (order of 10⁻³ /^oC), the terms 2α₁​α₂​T in the numerator and (α_1​+α_2​)T in denominator can be ignored.
 

Using the above approximation, 

${{\alpha }_p}_{ }\approx \frac{{\alpha }_1+{\alpha }_2}{2}$