1. Find all values of $p$, for each of which the numbers $|p-8|, 2p-1$, and $4p+5$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=-1, p=\frac{39}{8}$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 p-1)^{2}=(4 p+5)|p-8|$. For $p \geq 8$, we get $4 p^{2}-4 p+1=4 p^{2}-27 p-40$, from which $p=-\frac{41}{23}$ (does not fit). If $p<8$, then $4 p^{2}-4 p+1=-4 p^{2}+27 p+40$, $8 p^{2}-31 p-39=0$, hence $p=-1$ or $p=\frac{39}{8}$ (both roots fit).