# Sample Variance for MLE of Geometric Distribution

Given a random variable following the Geometric Distribution with parameter $p$ and so $f_X(\cdot) = p(1-p)^x$. The MLE of $p$ given a random sample of $n$ observations is $\hat{p}=\frac 1 {1+\bar{x}}$. How do I prove that the variance of $\bar{x}$ is $\frac{1-p}{np^2}$?

I have gone until the state where $\bar{x} = \frac{1-p}{p}$ but do not know how to continue from here...maybe, but this might be a far shot, $\frac 1{np} \bar{x}$?

How do I prove that the variance of $\bar{x}$ is $\frac{1-p}{np^2}$?
If $\bar{x}$ denotes the sample mean, then for any distribution (whose first two moments exist): $$Var(\bar{x}) = \frac{Var(X)}{n}$$ where $X$ is the underlying random variable. In your case, $X \sim Geometric(p)$ with $X = 0,1,2,\dots$, so simply calculate $Var(X) = \frac{(1 - p)}{p^2}$ and you are done.