For the terms of an arithmetic progression (AP), which of the following relations is always true?
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Answer and explanation
Correct answer: \(a_{17}=\frac{a_{12}+a_{22}}{2}\)
In an AP, \(a_n=a+(n-1)d\). Thus, \(a_{12}=a+11d\) and \(a_{22}=a+21d\); their average is \(a+16d=a_{17}\). In option B, the midpoint of the indices is 16.5, not 17. Exam tip: a term midway between two indices equals the average of those terms.
Frequently asked questions
What is the correct answer to this question?
\(a_{17}=\frac{a_{12}+a_{22}}{2}\)
Why is this the correct answer?
In an AP, \(a_n=a+(n-1)d\). Thus, \(a_{12}=a+11d\) and \(a_{22}=a+21d\); their average is \(a+16d=a_{17}\). In option B, the midpoint of the indices is 16.5, not 17. Exam tip: a term midway between two indices equals the average of those terms.
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.
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