13.146. Find three numbers, the first of which is as many times greater than the second as the second is greater than the third. If the first number is decreased by the sum of the other two, the result is 2, and if the first number is increased by half the difference of the second and the third, the result is 9.
Solution.

Let $x, y, z$ be the required numbers. According to the conditions $\left\{\begin{array}{l}\frac{x}{y}=\frac{y}{z}, \\ x-(y+z)=2, \\ x+\frac{y-z}{2}=9,\end{array}\right.$ $x=8, y=4, z=2$ or $x=-6.4 ; y=11.2 ; z=-19.6$.

Answer: $8 ; 4 ; 2$ or $-6.4 ; 11.2 ;-19.6$.