Mathematics Answer, Explanation and Revision Hints
यदि \(a_{n+1}-a_n=9\) और \(a_6=48\) है तो \(a_{20}\) क्या होगा? / If \(a_{n+1}-a_n=9\) and \(a_6=48\), what is \(a_{20}\)?
Correct Answer: C. (174). Explanation: यहां (d=9) है इसलिए \(a_{20}=48+14\times9=174\)। \(a_{n+1}-a_n\) समान्तर श्रेणी का सार्व अंतर होता है। / Here (d=9) so \(a_{20}=48+14\times9=174\). \(a_{n+1}-a_n\) is the common difference of the AP.
Which concept should I revise for this Mathematics MCQ?
Here (d=9) so \(a_{20}=48+14\times9=174\). \(a_{n+1}-a_n\) is the common difference of the AP.
What exam hint can help solve this Mathematics question?
यहां (d=9) है इसलिए \(a_{20}=48+14\times9=174\)। \(a_{n+1}-a_n\) समान्तर श्रेणी का सार्व अंतर होता है।
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