In an auditorium, the seats in rows are (40,48,56,\ldots). How many seats are there from the (30)th row to the (75)th row?

Correct answer: 20792

This is an AP with first term 40 and common difference 8. The 30th row has \(40+29\times8=272\) seats, and the 75th row has \(40+74\times8=632\) seats. There are \(75-30+1=46\) rows from the 30th to the 75th row. Therefore, the required total is \(\frac{46}{2}(272+632)=20792\). Taking only the sum up to the 75th row would incorrectly include the first 29 rows as well. Exam tip: for terms from one position to another, count them as \(\text{last position}-\text{first position}+1\).