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?

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 \(\text{last position}-\text{first position}+1\).

This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.