Overall effective activation energy for parallel reactions
The activation energy $E_\mathrm a$ should, in general, be viewed as an empirical quantity that characterises the sensitivity of the rate to temperature. From this perspective, it may be obtained via the equation

$$E_\mathrm a = RT^2 \left(\frac{\mathrm d \ln k}{\mathrm dT}\right) \tag{1}$$

This relation can be "derived" from the Arrhenius equation. (Quotation marks because the Arrhenius equation itself is only an empirical relation.)

$$\begin{align}
k &amp;= A\exp\left(-\frac{E_\mathrm a}{RT}\right) \\
\ln k &amp;= \ln A - \frac{E_\mathrm a}{RT} \\
\frac{\mathrm d \ln k}{\mathrm dT} &amp;= \frac{E_\mathrm a}{RT^2}
\end{align}$$

If a reaction rate has Arrhenius-type behaviour, i.e. $k = A\exp(-E_\mathrm a/RT)$, then equation $(1)$ simply returns the usual interpretation of the activation energy. However, equation $(1)$ is useful in that it allows us to define an "activation energy" in the case where the rate constant does not have the mathematical form $A\exp(-E/RT)$. The parallel reactions that you have described fall exactly into this category. The sum of two exponential functions generally does not return an exponential function, so the overall rate constant $k = k_1 + k_2$ should not be expected to conform to an Arrhenius description.

If we substitute

$$k = k_1 + k_2 = A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)$$

into equation $(1)$, then we get:

$$\begin{align}
E_\mathrm a &amp;= RT^2 \cdot \frac{\mathrm d}{\mathrm dT}\left\{ \ln\left[A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)\right] \right\} \\[8pt]
&amp;= RT^2 \cdot \frac{A_1(E_1/RT^2)\exp(-E_1/RT) + A_2(E_2/RT^2)\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&amp;= \frac{A_1E_1\exp(-E_1/RT) + A_2E_2\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&amp;= \frac{E_1k_1 + E_2k_2}{k_1 + k_2}
\end{align}$$

as desired.