When doing my research, I end up with examining the behavior of row-wise average of the following triangular array.
Let $A_n\stackrel{iid}{\sim} Bern(\pi)$, $Y_n\stackrel{iid}{\sim} N(\mu, \sigma^2)$, and $\pi_n=\dfrac{A_1+A_2+...+A_n}{n}$. Consider the following triangular array:
$\dfrac{A_1Y_1}{\pi_1}$
$\dfrac{A_1Y_1}{\pi_2}$, $\dfrac{A_2Y_2}{\pi_2}$
$\dfrac{A_1Y_1}{\pi_3}$, $\dfrac{A_2Y_2}{\pi_3}$, $\dfrac{A_3Y_3}{\pi_3}$
...
$\dfrac{A_1Y_1}{\pi_n}$, $\dfrac{A_2Y_2}{\pi_n}$, ..., $\dfrac{A_nY_n}{\pi_n}$
Let $Z_n$ is the average of row $n$. Does sequence $Z_n$ converge (a.s., in probability, or in distribution) to anything?
From the law of large numbers we have that, \begin{align} \pi_n \xrightarrow[]{p} \pi. \end{align} Similarly we have, \begin{align} \frac{1}{n}\sum_{i=1}^{n}A_iY_i \xrightarrow{p} \mathbb{E}[A_1Y_1]. \end{align} Now if $A_i$ and $Y_j$ are independent for all $i$ and $j$ then $\mathbb{E}[A_1Y_1] = \mathbb{E}[A_1]\mathbb{E}[Y_1] = \pi\mu$. This can also be interpreted as convergence in distribuiton, albeit with a degenerate distribution. Then, your situation fits with Slutsky's theorem and we have, \begin{align} \frac{\frac{1}{n}\sum_{i=1}^{n}A_iY_i}{\pi_n} \xrightarrow{d} \frac{\pi\mu}{\pi} = \mu. \end{align}
I believe that convergence in distribution towards a degenerate distribution, implies convergence in probability to the constant in the degenerate distribution. If you are concerned with dividing to zero, you can pad the numerator of $\pi_n$ with an arbitrarily small value $\delta>0$.