2.1. The numbers 7, 8, 9, 10, 11 are arranged in a row in some order. It turned out that the sum of the first three of them is 26, and the sum of the last three is 30. Determine the number standing in the middle.

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Answer: 11

Solution. If we add the sum of the first three and the sum of the last three, we get the sum of all four numbers once and the middle number twice. Since the sum of all five numbers is $7+8+9+10+11=45$, and the sum of 26 and 30 is 56, then the number counted twice is 11.