Two identical metal wires of thermal conductivities $$K_1$$ and $$K_2$$ respectively are connected in series. The effective thermal conductivity of the combination is:

For two identical wires (same length $$l$$ and cross-sectional area $$A$$) connected in series, the total thermal resistance is the sum of individual resistances. The thermal resistance of each wire is $$R_i = \frac{l}{K_i A}$$, so the total resistance is $$R = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

The effective conductivity $$K_{\text{eff}}$$ of the combination (total length $$2l$$, same area $$A$$) satisfies $$R = \frac{2l}{K_{\text{eff}} A}$$. Therefore $$\frac{2l}{K_{\text{eff}} A} = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

Simplifying: $$\frac{2}{K_{\text{eff}}} = \frac{1}{K_1} + \frac{1}{K_2} = \frac{K_1 + K_2}{K_1 K_2}$$, which gives $$K_{\text{eff}} = \frac{2K_1 K_2}{K_1 + K_2}$$.

The correct answer is option 1: $$\frac{2K_1 K_2}{K_1 + K_2}$$.