$10\mathrm{.}8.$ Let $x[n]$ be a signal whose rational $z-$transform $x(z)$ contains a pole at $z=\frac{1}{2}.$ Given that $x_1[n]=(\frac{1}{4}{)}^{n}x[n]$ is absolutely summable and $x_2[n]=(\frac{1}{8}{)}^{n}x[n]$