3. (7 points) Given an arithmetic progression. The sum of its first 10 terms is 60, and the sum of its first 20 terms is 320. What can the 15th term of this progression be?
Answer: 25.

Solution. Let the first term of the sequence be $a$, and the common difference be $b$. Then the sum of the first 10 terms is $a+(a+b)+\ldots+(a+9b)=$ $10a+45b$. The sum of the first 20 terms is $a+(a+b)+\ldots+(a+19b)=$ $20a+190b$. According to the problem, $10a+45b=60, 20a+190b=320$. Solving the system, we find $a=-3, b=2$. Then the 15th term is $a+14b=25$.

Criteria. Any correct solution: 7 points.

A generally correct solution containing arithmetic errors that do not affect the solution process: 5 points.

Only the answer: 1 point.