10.5. An increasing geometric progression consists of four different positive numbers, three of which form an arithmetic progression. What can the denominator of this progression be? Provide all possible answers and prove that there are no others.
Solution: Consider an arbitrary increasing geometric progression of four numbers $b, b q, b q^{2}, b q^{3} (b>0, q>1)$. If the first three terms (or the last three) form an arithmetic progression, then we have the equation $2 b q=b+b q^{2}$, from which $q^{2}-2 q+1=0 \Leftrightarrow q=1$. This is a contradiction.

Suppose the arithmetic progression is formed by the first, second, and fourth terms. This condition is equivalent to the equation

$$
2 b q=b+b q^{3} \Leftrightarrow q^{3}-2 q+1=0 \Leftrightarrow(q-1)\left(q^{2}+q-1\right)=0
$$

The obtained equation has three roots. $q_{1}=1, q_{2}=\frac{-1-\sqrt{5}}{2}$, and $q_{3}=\frac{-1+\sqrt{5}}{2}$. However, none of them satisfy the condition $q>1$. This case is also impossible.

Suppose the arithmetic progression is formed by the first, third, and fourth terms. This condition is equivalent to the equation

$$
2 b q^{2}=b+b q^{3} \Leftrightarrow q^{3}-2 q^{2}+1=0 \Leftrightarrow(q-1)\left(q^{2}-q-1\right)=0
$$

The obtained equation has three roots. $q_{1}=1$ and $q_{2}=\frac{1-\sqrt{5}}{2}$ do not satisfy the condition $q>1$. But the third root $q_{3}=\frac{1+\sqrt{5}}{2}$ does satisfy it. In this case, the number $b$ can be any positive number - the condition of the problem will be satisfied.

Answer: $\frac{1+\sqrt{5}}{2}$.

Note 1: We can immediately assume that $b$ is any positive number, for example, 1, since it is obvious that when all terms of the sequence are reduced (increased) by the same factor, the property of being a progression (whether geometric or arithmetic) is preserved, and the common ratio of the geometric progression remains the same. This fact should be considered obvious when checking the solution.

Note 2: In the provided solution, the last situation can be easily reduced to the previous one by replacing the increasing progression with a decreasing one and reversing the order of the numbers. Of course, with such a replacement, we will be interested in $q$ from the interval $(0 ; 1)$, and the answer will be the number equal to $\frac{1}{q}$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Correctly analyzed two out of three situations:  1) three consecutive terms of the geometric progression form an arithmetic progression;  2) the first, second, and fourth terms form an arithmetic progression;  3) the first, third, and fourth terms form an arithmetic progression | 5 points |
| Correctly analyzed situation 1 (see 5-point item) and provided a correct example of four numbers (with the common ratio of the progression $\frac{1+\sqrt{5}}{2}$) | 4 points |
| Correctly analyzed only one out of three situations (see 5-point item) | 3 points |
| Provided a correct example of four numbers (with the common ratio of the progression $\frac{1+\sqrt{5}}{2}$) | 1 point |
| Answer without justification and/or incorrect answer | 0 points |