1. Find all values of $p$, for each of which the numbers $9 p+10, 3 p$ and $|p-8|$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=-1, p=\frac{40}{9}$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 p)^{2}=(9 p+10) p-8 \mid$. For $p \geq 8$, we get $9 p^{2}=9 p^{2}-62 p-80$, from which $p=-\frac{40}{31} \quad$ (does not fit). If $p<8$, then $9 p^{2}=-9 p^{2}+62 p+80$, $9 p^{2}-31 p-40=0$, hence, $p=-1$ or $p=\frac{40}{9}$ (both roots fit).