यदि समांतर श्रेणी में \(a_4+a_{12}=112\) है, तो \(a_8\) क्या होगा?
If in an arithmetic progression \(a_4+a_{12}=112\), what is \(a_8\)?
Explanation opens after your attempt
Correct Answer
C. (56)
Concept
\(a_8\) is equidistant from \(a_4\) and \(a_{12}\), so \(a_8=\frac{112}{2}=56\). The average of symmetric terms is the middle term.
Why this answer is correct
The correct answer is C. (56). \(a_8\) is equidistant from \(a_4\) and \(a_{12}\), so \(a_8=\frac{112}{2}=56\). The average of symmetric terms is the middle term.
Exam Tip
\(a_8\), \(a_4\) और \(a_{12}\) से समान दूरी पर है, इसलिए \(a_8=\frac{112}{2}=56\)। सममित पदों का औसत मध्य पद होता है।
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