11.3. The numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ form a geometric progression. Among them, there are both rational and irrational numbers. What is the maximum number of terms in this progression that can be rational numbers?
Answer: 3.

Example: let $a_{1}=1, q=\sqrt{2}$, we get the geometric progression $1, \sqrt{2}, 2, 2\sqrt{2}, 4$. Evaluation. If there are 4 rational numbers among them, then there will be two consecutive rational members of the geometric progression. This means that the common ratio of the progression (the ratio of the subsequent term to the previous one) is a rational number. But then, with a rational $a_{1}$, all terms of the progression are rational numbers, and with an irrational $a_{1}$, all terms are irrational.

Remarks. An example of a geometric progression with three rational and two irrational numbers is given - 3 points. It is proven that there cannot be more than three rational numbers, but no example is provided - 3 points. If both are present, 7 points.