गुणोत्तर श्रेणी \(2,8,32,\ldots\) में (2048) तक कितने पद हैं?

How many terms are there up to (2048) in the geometric progression \(2,8,32,\ldots\)?

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Correct Answer

B. (6)

Step 1

Concept

\(2\cdot4^{n-1}=2048\), so \(4^{n-1}=1024=4^5\) and (n=6). In exams, equate the last term to the general term.

Step 2

Why this answer is correct

The correct answer is B. (6). \(2\cdot4^{n-1}=2048\), so \(4^{n-1}=1024=4^5\) and (n=6). In exams, equate the last term to the general term.

Step 3

Exam Tip

\(2\cdot4^{n-1}=2048\) से \(4^{n-1}=1024=4^5\), इसलिए (n=6) है। परीक्षा में अंतिम पद को सामान्य पद के बराबर रखें।

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Mathematics Answer, Explanation and Revision Hints

गुणोत्तर श्रेणी \(2,8,32,\ldots\) में (2048) तक कितने पद हैं? / How many terms are there up to (2048) in the geometric progression \(2,8,32,\ldots\)?

Correct Answer: B. (6). Explanation: \(2\cdot4^{n-1}=2048\) से \(4^{n-1}=1024=4^5\), इसलिए (n=6) है। परीक्षा में अंतिम पद को सामान्य पद के बराबर रखें। / \(2\cdot4^{n-1}=2048\), so \(4^{n-1}=1024=4^5\) and (n=6). In exams, equate the last term to the general term.

Which concept should I revise for this Mathematics MCQ?

\(2\cdot4^{n-1}=2048\), so \(4^{n-1}=1024=4^5\) and (n=6). In exams, equate the last term to the general term.

What exam hint can help solve this Mathematics question?

\(2\cdot4^{n-1}=2048\) से \(4^{n-1}=1024=4^5\), इसलिए (n=6) है। परीक्षा में अंतिम पद को सामान्य पद के बराबर रखें।