2. The sum of the members of an infinite decreasing geometric progression that stand at odd positions is 36, and the sum of the members that stand at even positions of the same geometric progression is 12. Find the progression.
Solution. If $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is an infinite decreasing geometric progression with ratio $q$, then the sequences $a_{1}, a_{3}, \ldots, a_{2 k+1}, \ldots$ and $a_{2}, a_{4}, \ldots, a_{2 k}, \ldots$, are infinite geometric progressions with ratio $q^{2}$. According to the condition of the problem, the last two sequences have sums of 36 and 12 respectively, i.e.

$$
\frac{a_{1}}{1-q^{2}}=36 \quad \text { and } \quad \frac{a_{2}}{1-q^{2}}=12 .
$$

Considering that $a_{2}=a_{1} q$, solving the system

$$
\frac{a_{1}}{1-q^{2}}=36, \frac{a_{1} q}{1-q^{2}}=12
$$

we get $a_{1}=32, q=\frac{1}{3}$.

Thus, the required progression is: $32, \frac{32}{3}, \frac{32}{9}, \ldots$.