10.2. Let $A$ be a set of ten distinct positive numbers (not necessarily integers). Determine the maximum possible number of arithmetic progressions consisting of three distinct numbers from the set $A$.
Answer: 20.

Solution: Let the elements of set $A$ be denoted as $a_{1}<a_{2}<\ldots<a_{10}$. Three numbers $a_{k}<a_{l}<a_{m}$ form a three-term arithmetic progression if and only if $a_{l}-a_{k}=a_{m}-a_{l}$. Let's see how many times each element of $A$ can be the middle term $a_{l}$ of such a progression. It is easy to see that for an element $a_{l}, l=2,3,4,5$, the first element $a_{k}$ can be chosen in no more than $l-1$ different ways - it can only be $a_{1}, a_{2}, \ldots, a_{l-1}$. Therefore, the number of such progressions with the middle term $a_{l}, l=2,3,4,5$ cannot exceed $1+2+3+4=10$. Similarly, for an element $a_{l}, l=6,7,8,9$, the third element $a_{m}$ can be chosen in no more than $10-l$ different ways - it can only be $a_{l+1}, a_{l+2}, \ldots, a_{10}$. Therefore, the number of such progressions with the middle term $a_{l}, l=6,7,8,9$ cannot exceed $4+3+2+1=10$. Any sought three-term progression must be one of these two types, so the total number of such ways cannot exceed $1+2+3+4+4+3+2+1=20$.

As an example of a set $A$ with 10 elements where the value 20 is achieved, one can take the set of all natural numbers from 1 to 10 inclusive.

Grading criteria. ($\cdot$) Proven that the number of progressions is no more than 20: 5 points. ($\cdot$) Example with 20 progressions: 2 points.