1B. Let A be the sum of the terms of a finite geometric progression (the terms are positive) and $B$ be the sum of their reciprocals. Find the product of the terms of the progression.
Solution. Let the members of the geometric progression be $a_{1}, a_{2}, \ldots, a_{n}$. If we denote the ratio of the progression by $q$, then

$$
P=a_{1} a_{2} \ldots a_{n}=a_{1}^{n} q^{\frac{(n-1) n}{2}}=\left(a_{1}^{2} q^{n-1}\right)^{\frac{n}{2}}=\left(a_{1} a_{n}\right)^{\frac{n}{2}}
$$

On the other hand, it holds that

$a_{i} a_{n+1-i}=a_{1} q^{i-1} a_{1} q^{n-i}=a_{1} a_{1} q^{n-1}=a_{1} a_{n}$,

for every $i \in\{1,2, \ldots, n\}$. Then

$$
A=a_{1}+a_{2}+\ldots+a_{n}=\frac{a_{1} a_{n}}{a_{n}}+\frac{a_{2} a_{n-1}}{a_{n-1}}+\ldots+\frac{a_{n} a_{1}}{a_{1}}=a_{1} a_{n}\left(\frac{1}{a_{n}}+\frac{1}{a_{n-1}}+\ldots+\frac{1}{a_{1}}\right)=a_{1} a_{n} B
$$

Thus $a_{1} a_{n}=\frac{A}{B}$, and therefore $P=\left(\frac{A}{B}\right)^{\frac{n}{2}}$.