1. Find all values of $p$, for each of which the numbers $-p-12, 2 \cdot \sqrt{p}$, and $p-5$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=4$.

Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(-p-12)(p-5)$, from which

$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+11 p-60=0
\end{array} \Leftrightarrow p=4\right.
$$