In an AP, (a_p=42), (a_{p+12}=150), and (p=9). What is (a_{40})?
Answer and explanation
Correct answer: 321
Since \(p=9\), we have \(a_9=42\) and \(a_{21}=150\). Thus, \(a_{21}-a_9=12d=150-42=108\), so \(d=9\). Now \(a_{40}=a_9+(40-9)d=42+31\times9=321\). The value 330 does not follow from the common difference \(d=9\) for the 40th term. Exam tip: When two terms with different indices are given, first use the difference of their indices to find \(d\).
Frequently asked questions
What is the correct answer to this question?
321
Why is this the correct answer?
Since \(p=9\), we have \(a_9=42\) and \(a_{21}=150\). Thus, \(a_{21}-a_9=12d=150-42=108\), so \(d=9\). Now \(a_{40}=a_9+(40-9)d=42+31\times9=321\). The value 330 does not follow from the common difference \(d=9\) for the 40th term. Exam tip: When two terms with different indices are given, first use the difference of their indices to find \(d\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
