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?
Answer and explanation
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.
Frequently asked questions
What is the correct answer to this question?
4250
Why is this the correct answer?
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.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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