In an AP, (a_3+a_8+a_{13}=156). Find the value of (a_8).
Answer and explanation
Correct answer: 52
In an AP, terms with equally spaced indices have the middle term as their average. Hence, \(a_3+a_{13}=2a_8\). Therefore, \(a_3+a_8+a_{13}=3a_8=156\), giving \(a_8=52\). If \(a_8=54\), the sum would be 162, so it cannot be correct. Exam tip: For \(a_{n-r}, a_n, a_{n+r}\) in an AP, use \(a_{n-r}+a_{n+r}=2a_n\).
Frequently asked questions
What is the correct answer to this question?
52
Why is this the correct answer?
In an AP, terms with equally spaced indices have the middle term as their average. Hence, \(a_3+a_{13}=2a_8\). Therefore, \(a_3+a_8+a_{13}=3a_8=156\), giving \(a_8=52\). If \(a_8=54\), the sum would be 162, so it cannot be correct. Exam tip: For \(a_{n-r}, a_n, a_{n+r}\) in an AP, use \(a_{n-r}+a_{n+r}=2a_n\).
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.
