3B. In the geometric progression $a_{1}, a_{2}, \ldots$ the terms $A=a_{m+n}$ and $B=a_{m-n}$ are given. Find $a_{m}$ and $a_{n}$.
Solution. Let $q$ be the common ratio of the progression. Then $A=a_{m+n}=a_{1} q^{m+n-1}$ and $B=a_{m-n}=a_{1} q^{m-n-1}$. From this, $q^{2 n}=\frac{A}{B}$, so $q=\sqrt[2 n]{\frac{A}{B}}$. Therefore,

$$
\begin{aligned}
& a_{m}=a_{1} q^{m-1}=a_{1} q^{m-n-1} q^{n}=a_{m-n} q^{n}=B\left(\sqrt[2 n]{\frac{A}{B}}\right)^{n}=\sqrt{A B} \text { and } \\
& a_{n}=a_{1} q^{n-1}=a_{1} q^{m+n-1} q^{-m}=a_{m+n} q^{-m}=A\left(\frac{A}{B}\right)^{-\frac{m}{2 n}}=A^{1-\frac{m}{2 n}} B^{\frac{m}{2 n}}
\end{aligned}
$$