यदि \(a_{n+1}-a_n=8\) और \(a_4=30\) है, तो \(a_{16}\) क्या होगा?

If \(a_{n+1}-a_n=8\) and \(a_4=30\), what is \(a_{16}\)?

Explanation opens after your attempt
Correct Answer

C. (126)

Step 1

Concept

Here (d=8), so \(a_{16}=a_4+12d=30+96=126\). \(a_{n+1}-a_n\) is the common difference of the AP.

Step 2

Why this answer is correct

The correct answer is C. (126). Here (d=8), so \(a_{16}=a_4+12d=30+96=126\). \(a_{n+1}-a_n\) is the common difference of the AP.

Step 3

Exam Tip

यहां (d=8), इसलिए \(a_{16}=a_4+12d=30+96=126\)। \(a_{n+1}-a_n\) ही AP 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 \(a_{n+1}-a_n=8\) और \(a_4=30\) है, तो \(a_{16}\) क्या होगा? / If \(a_{n+1}-a_n=8\) and \(a_4=30\), what is \(a_{16}\)?

Correct Answer: C. (126). Explanation: यहां (d=8), इसलिए \(a_{16}=a_4+12d=30+96=126\)। \(a_{n+1}-a_n\) ही AP का सार्व अंतर होता है। / Here (d=8), so \(a_{16}=a_4+12d=30+96=126\). \(a_{n+1}-a_n\) is the common difference of the AP.

Which concept should I revise for this Mathematics MCQ?

Here (d=8), so \(a_{16}=a_4+12d=30+96=126\). \(a_{n+1}-a_n\) is the common difference of the AP.

What exam hint can help solve this Mathematics question?

यहां (d=8), इसलिए \(a_{16}=a_4+12d=30+96=126\)। \(a_{n+1}-a_n\) ही AP का सार्व अंतर होता है।