$13 \cdot 13$ Given 4 weights forming a geometric progression and a two-pan balance, explain how to find the heaviest weight using the balance only twice.

Given 4 weights forming a geometric progression and a two-pan balance, explain how to find the heaviest weight using the balance only twice.
［Solution］Assume the weights of the weights are
$$
a, a r, a r^{2}, a r^{3} \quad(r>1) . \cdot
$$

First weighing: Place two weights on each pan of the balance. Since $r>1$, then
$$
\begin{array}{l}
\left(a r^{3}+a\right)-\left(a r^{2}+a r\right)=a\left(r^{2}-1\right)(r-1)>0, \\
\left(a r^{3}+a r\right)-\left(a r^{2}+a\right)=a\left(r^{2}+1\right)(r-1)>0, \\
\left(a r^{3}+a r^{2}\right)-(a r+a)=a\left(r^{2}-1\right)(r+1)>0,
\end{array}
$$

Therefore, the pan $(H)$ containing $a r^{3}$ will be heavier.
Second weighing: Just compare the two weights on the heavier pan $(H)$ to find the heaviest weight $a r^{3}$.