The Expected Number of Bernoulli Trials Needed for $n$ successes in a row for any $p$ was answered by a member on math.stackexchange, and also found in this writeup on page two: https://courses.cit.cornell.edu/info2950_2012sp/mh.pdf

Let $A_n$ be the expected/average/mean number of coin flips/bernoulli trials with probability of success $p$ to get $n$ number of heads in a row.

$large A_n=frac{p^{−n}−1}{1−p}$

Do the Binomial Distribution formulas for the mean, variance and standard deviation work for this An formula that was found?

Can I assume the variance is just $σ^2=μ(1−p)$ with $μ=A_n$?

If yes, then the standard deviation is $σ=sqrt{μ(1−p)}$ with $μ=A_n$

I am not sure it is this straightforward because our expression for the mean is more complex than just np.