1. Find all values of $p$, for each of which the numbers $4 p+5, 2 p$ and $|p-3|$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=-1, p=\frac{15}{8}$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 p)^{2}=(4 p+5)|p-3|$. For $p \geq 3$, we get $4 p^{2}=4 p^{2}-7 p-15$, from which $p=-\frac{15}{7}$ (does not fit). If $p<3$, then $4 p^{2}=-4 p^{2}+7 p+15, 8 p^{2}-7 p-15=0$, hence $p=-1$ or $p=\frac{15}{8}$ (both roots fit).