2. (2 points) The sequence $x_{n}$ is defined by the following conditions: $x_{1}=2, x_{1}+x_{2}+\ldots+$ $x_{n-1}+\frac{3}{2} x_{n}=3$. Find $x_{1000}$.
Answer: $x_{1000}=\frac{2}{3^{999}}$.

Solution:

Notice that the equality $x_{1}+x_{2}+\ldots+x_{n-1}+\frac{3}{2} x_{n}=3$ for $n=1$ also holds: $\frac{3}{2} x_{1}=3$.

For $n \geqslant 2$, by subtracting the equality $x_{1}+x_{2}+\ldots+x_{n-1}+\frac{3}{2} x_{n}=2$ from the equality $x_{1}+x_{2}+\ldots+\frac{3}{2} x_{n-1}=3$, we get that $\frac{3}{2} x_{n}-\frac{1}{2} x_{n-1}=0$, from which $x_{n}=\frac{1}{3} x_{n-1}$ for $n \geqslant 2$.

Thus, $x_{n}$ is a geometric progression with the first term 2 and the common ratio $\frac{1}{3}$, so $x_{1000}=\frac{2}{3^{999}}$.