Show estimator is consistent

For a special case of the gamma distribution:

$$f(x)=frac{x}{theta^2}e^{-x/ theta}$$

$$E(x) = 2theta\ V(x) = 2theta^2$$

I find MLE of $hat{theta}$ to be $sum_{i=1}^n frac{x_i}{2n}$

To show consistency, I’d like to show that:

• $lim_{nto infty} E(hat{theta}) = theta$
• $lim_{nto infty} V(hat{theta}) = 0$

Focusing on $lim_{nto infty} V(hat{theta}) = 0$:

$V(sum_{i=1}^n frac{x_i}{2n}) = frac{1}{4n^2}V(sum_{i=1}^n x_i)$

I’m somewhat stuck here, it seems obvious that this limit approaches, $0$, however, I’m not sure if I need to break the variance further (say $V(x) = E(x^2)-E(x)^2$) to prove this

I’d appreciate any help!