In an auditorium, the seats in rows are (20,24,28,\ldots). How many seats are there from the (15)th row to the (35)th row?

2436

This is an AP with first term \(a=20\) and common difference \(d=4\). The 15th row has \(20+14\times4=76\) seats, while the 35th row has \(20+34\times4=156\) seats. There are \(35-15+1=21\) rows from the 15th to the 35th row, inclusive. Hence, the required sum is \(\frac{21}{2}(76+156)=2436\). Option 2408 may result from counting the rows or finding the last term incorrectly. Exam tip: When summing from one term number to another, include both endpoints by adding \(+1\) to the difference of their positions.

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