1. Find all values of $p$, for each of which the numbers $-p-8$, $3 \cdot \sqrt{p}$, and $p-7$ are the first, second, and third terms, respectively, 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 $(3 \sqrt{p})^{2}=(-p-8)(p-7)$, from which

$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+10 p-56=0
\end{array} \Leftrightarrow p=4\right.
$$