When using arithmetic returns the right way to calculate an average is via the geometric average. The reason is that there is a multiplicative relationship between the returns. Example: Let $P_t$ denote the stock price at time $t$, then the simple (arithmetic) net return is defined as:
\begin{equation}
r_t=\frac{P_t-P_{t-1}}{P_{t-1}}=\frac{P_t}{P_{t-1}}-1
\end{equation}
Now look at:
\begin{align}
r_t[k]&=\frac{P_t}{P_{t-k}}-1=\frac{P_t}{P_{t-1}}\cdot \frac{P_{t-1}}{P_{t-2}} \dots \frac{P_{t-k+1}}{P_{t-k}}-1=(1+r_t)\cdot(1+r_{t-1})\dots(1+r_{t-k+1})-1\\&=\prod_{j=0}^{k-1}(1+r_{t-j})-1
\end{align}
The annualized average return (geometric average of $k-1$ period returns) is given by
\begin{align}
\overline{r_t[k]}&=\left(\prod_{j=0}^{k-1}(1+r_{t-j})\right)^{\frac{1}{k}}-1\\&=\exp\left(\ln\left(\prod_{j=0}^{k-1}(1+r_{t-j})\right)^\frac{1}{k} \right)-1\\&=\exp\left(\frac{1}{k} \sum_{j=0}^{k-1}\ln(1+r_{t-j}) \right)-1
\end{align}
However, notice that this expression is quite difficult to compute. So it is often approximated by the arithmetic average of the $k-1$-period returns
\begin{equation}
\overline{r_t[k]} \approx \frac{1}{k}\sum_{j=0}^{k-1}r_{t-j}
\end{equation}
This linear approximation obtains from a 1-order Taylor expansion around zero. So if the returns $r_{t-j}$ are small you get almost the same result with less effort. Especially when looking at daily returns this approximation is very good in most cases, because $r_{j-i} \approx 0$.
When you are using log-returns it is even simpler, because there is no multiplicative relationship and the "right" average is the simple average.
Regards.