1. Find all values of $p$, for each of which the numbers $|p-3|, 3p+1$, and $9p+10$ are respectively the first, second, and third terms of some geometric progression.
Answer: $p=-1, p=\frac{29}{18}$.

Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 p+1)^{2}=(9 p+10)|p-3|$. For $p \geq 3$, we get $9 p^{2}+6 p+1=9 p^{2}-17 p-30$, from which $p=-\frac{31}{23}$ (does not fit). If $p<3$, then $9 p^{2}+6 p+1=-9 p^{2}+17 p+30$, $18 p^{2}-11 p-29=0$, hence $p=-1$ or $p=\frac{29}{18}$ (both roots fit).