Medium Mathematics Arithmetic Progressions (AP) Class 10 Level 65

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 \(a_{20}\)?

Explanation opens after your attempt
Correct Answer

C. (174)

Step 1

Concept

Here (d=9) so \(a_{20}=48+14\times9=174\). \(a_{n+1}-a_n\) is the common difference of the AP.

Step 2

Why this answer is correct

The correct answer is C. (174). Here (d=9) so \(a_{20}=48+14\times9=174\). \(a_{n+1}-a_n\) is the common difference of the AP.

Step 3

Exam Tip

यहां (d=9) है इसलिए \(a_{20}=48+14\times9=174\)। \(a_{n+1}-a_n\) समान्तर श्रेणी का सार्व अंतर होता है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

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।

Student Class Required

Select your class first

Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.