1. Find all values of $p$, for each of which the numbers $p-2$, $2 \cdot \sqrt{p}$, and $-3-p$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=1$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(p-2)(-p-3)$, from which

$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+5 p-6=0
\end{array} \Leftrightarrow p=1\right.
$$