47. Can the numbers $2, \sqrt{6}$, and 4.5 be members of the same arithmetic or the same geometric progression?
47. Let $2, \sqrt{6}$ and 4.5 be terms of one arithmetic progression, then 72
$\sqrt{6}=2+m d, 4.5=2+n d$. In the first case $d=\frac{\sqrt{6}-2}{m}$ is an irrational number. In the second case $d=\frac{5}{2 n}$ is a rational number. The numbers $2, \sqrt{6}$ and 4.5 are not terms of one arithmetic progression.

Let $2, \sqrt{6}$ and 4.5 be terms of one geometric progression, then $\sqrt{6}=2 q^{m}$, $4.5=2 q^{n}$. Hence $q^{m}=\sqrt{\frac{3}{2}}, q^{n}=\frac{9}{4}$,

$$
q^{m n}=\left(\sqrt{\frac{3}{2}}\right)^{n} ; \quad q^{m n}=\left(\frac{9}{4}\right)^{m} \text { or }\left(\frac{3}{2}\right)^{\frac{n}{2}}=\left(\frac{3}{2}\right)^{2 m}, n=4 m
$$

The numbers $2, \sqrt{6}$ and 4.5 can be terms of one geometric progression when $q=\sqrt{\frac{3}{2}}$.