Let the length of each wire be $l$ and the area of cross-section be $A$. The resistance of a wire is related to its conductivity $\sigma$ by the formula $R = \frac{l}{\sigma A}$. For the first wire, $R_1 = \frac{l}{\sigma_1 A}$. For the second wire, $R_2 = \frac{l}{\sigma_2 A}$. When the wires are connected in series, their equivalent resistance $R_{eq}$ is the sum of their individual resistances: $R_{eq} = R_1 + R_2 = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right) = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$. For the equivalent composite wire, the total length is $l_{eq} = l + l = 2l$, and the area of cross-section remains the same ($A_{eq} = A$). Let its effective conductivity be $\sigma_{eq}$. The equivalent resistance can also be written as: $R_{eq} = \frac{l_{eq}}{\sigma_{eq} A_{eq}} = \frac{2l}{\sigma_{eq} A}$. Equating the two expressions for $R_{eq}$: $\frac{2l}{\sigma_{eq} A} = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$ $\frac{2}{\sigma_{eq}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$ $\sigma_{eq} = \frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$.