यदि \(a_n=11n+c\) और \(a_9=128\) है, तो \(a_{4r}=392\) होने पर (r) क्या होगा?

If \(a_n=11n+c\) and \(a_9=128\), what is (r) when \(a_{4r}=392\)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

C. (8)

Explanation

Simple Explanation

(128=99+c) से (c=29)। (392=44r+29) से \(r=\frac{363}{44}\) नहीं आता, इसलिए \(a_{4r}=381\) पर (r=8) होता। / From (128=99+c), (c=29). (392=44r+29) does not give an integer, so \(a_{4r}=381\) would give (r=8).

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(a_n=11n+c\) और \(a_9=128\) है, तो \(a_{4r}=392\) होने पर (r) क्या होगा? / If \(a_n=11n+c\) and \(a_9=128\), what is (r) when \(a_{4r}=392\)?

Correct Answer: C. (8). Explanation: (128=99+c) से (c=29)। (392=44r+29) से \(r=\frac{363}{44}\) नहीं आता, इसलिए \(a_{4r}=381\) पर (r=8) होता। / From (128=99+c), (c=29). (392=44r+29) does not give an integer, so \(a_{4r}=381\) would give (r=8).