गुणोत्तर श्रेणी \(12,48,192,768,\ldots\) का सामान्य पद क्या है?
What is the general term of the geometric progression \(12,48,192,768,\ldots\)?
Explanation opens after your attempt
A. \(a_n=12\cdot4^{n-1}\)
Concept
The first term is (12) and the ratio is (4), so \(a_n=12\cdot4^{n-1}\). In exams, keep (a) and (r) correct in \(ar^{n-1}\).
Why this answer is correct
The correct answer is A. \(a_n=12\cdot4^{n-1}\). The first term is (12) and the ratio is (4), so \(a_n=12\cdot4^{n-1}\). In exams, keep (a) and (r) correct in \(ar^{n-1}\).
Exam Tip
पहला पद (12) और अनुपात (4) है, इसलिए \(a_n=12\cdot4^{n-1}\) है। परीक्षा में \(ar^{n-1}\) में (a) और (r) सही रखें।
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