134. Three numbers are consecutive terms of some geometric progression. If 2 is added to the second number, they will form an arithmetic progression. If after that 16 is added to the third number, they will again form a geometric progression. Find these numbers.
Solution. Let $b_{1}, b_{2}, b_{3}$ be consecutive terms of a geometric progression. Then, by the condition, the numbers $b_{1}, b_{2}+2, b_{3}$ form an arithmetic progression, and the numbers $b_{1}, b_{2}+2, b_{3}+16$ form a geometric progression. Based on the properties of arithmetic and geometric progressions, we can set up the following system of equations:

$$
\left\{\begin{array}{l}
b_{2}^{2}=b_{1} b_{3} \\
b_{2}+2=\frac{b_{1}+b_{3}}{2} \\
\left(b_{2}+2\right)^{2}=b_{1}\left(b_{3}+16\right)
\end{array}\right.
$$

Solving this system, we find $b_{1}=1$ or $b_{1}=\frac{1}{9}, b_{2}=3$ or $b_{2}=-\frac{5}{9}$, $b_{3}=9$ or $b_{3}=\frac{25}{9}$.

Answer: $1, 3, 9$ or $\frac{1}{9}, -\frac{5}{9}, \frac{25}{9}$.