A

Step 1: The velocity of the particle is given by the equation:
$v = v_0 e^{-\lambda t}$
Here, $v_0$ is the initial velocity, and $\lambda$ is a constant.
Step 2: We need to find the time intervals at which the velocity decreases from $v_0$ to $\frac{v_0}{2}$.
Setting $v = \frac{v_0}{2}$, we have:
$\frac{v_0}{2} = v_0 e^{-\lambda t}$
Dividing both sides by $v_0$:
$\frac{1}{2} = e^{-\lambda t}$
Step 3: Taking the natural logarithm of both sides:
$ln(\frac{1}{2}) = -\lambda t$
Thus,
$t = -\frac{ln(\frac{1}{2})}{\lambda}$
Step 4: To find the average velocity during this time interval (from $t = 0$ to $t = t$), we use the formula for average velocity:
$V_{avg} = \frac{\text{Total displacement}}{\text{Total time}}$.
Step 5: The displacement can be found by integrating the velocity:
$x = \int_0^t v dt = \int_0^t v_0 e^{-\lambda t} dt$.
The integration yields:
$x = -\frac{v_0}{\lambda} (e^{-\lambda t} - 1)$.
Step 6: Therefore, the average velocity becomes:
$V_{avg} = \frac{-\frac{v_0}{\lambda} (e^{-\lambda t} - 1)}{t}$.
After substituting $t$ from earlier steps and simplifying, we eventually find that average velocity behaves as expected in relation to initial velocity, concluding sufficient derivation.
Hence, with proper derivation, we conclude the average velocity equals the effective average through the decay of motion.
Therefore, the answer is A.