यदि समान्तर श्रेणी में \(a_7=7x-8\), \(a_{15}=15x-48\) और \(a_{23}=23x-88\) हैं, तो \(a_{31}\) क्या होगा?
If in an AP \(a_7=7x-8\), \(a_{15}=15x-48\), and \(a_{23}=23x-88\), what is \(a_{31}\)?
Explanation opens after your attempt
C. (31x-128)
Concept
The group difference of equally spaced terms is (8x-40). \(a_{31}=a_{23}+8x-40=31x-128\).
Why this answer is correct
The correct answer is C. (31x-128). The group difference of equally spaced terms is (8x-40). \(a_{31}=a_{23}+8x-40=31x-128\).
Exam Tip
समान दूरी वाले पदों का समूह अंतर (8x-40) है। \(a_{31}=a_{23}+8x-40=31x-128\)।
Login to save your score, XP, coins and progress.
