In an auditorium, the seats in rows are (25,30,35,\ldots). How many seats are there from the (18)th row to the (42)nd row?

Correct answer: 4250

This is an AP with first term \(a=25\) and common difference \(d=5\). The 18th row has \(a_{18}=25+17\times5=110\) seats, and the 42nd row has \(a_{42}=25+41\times5=230\) seats. There are \(42-18+1=25\) rows from the 18th through the 42nd row. Hence, the sum is \(\frac{25}{2}(110+230)=4250\). An option such as 4175 can result from incorrectly counting the rows without including both end rows. Exam tip: for an inclusive range, use \(\text{last index}-\text{first index}+1\) for the number of terms.