गुणोत्तर श्रेणी \(729,243,81,27,\ldots\) में (1) कौन-सा पद है?

In the geometric progression \(729,243,81,27,\ldots\), which term is (1)?

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Correct Answer

A. सातवाँ पद(7)th term

Step 1

Concept

From (729\left\(\frac{1}{3}\right\)^{n-1}=1), \(3^{6-(n-1)}=1\), so (n=7). In exams count decreasing powers carefully.

Step 2

Why this answer is correct

The correct answer is A. सातवाँ पद / (7)th term. From (729\left\(\frac{1}{3}\right\)^{n-1}=1), \(3^{6-(n-1)}=1\), so (n=7). In exams count decreasing powers carefully.

Step 3

Exam Tip

(729\left\(\frac{1}{3}\right\)^{n-1}=1) से \(3^{6-(n-1)}=1\) इसलिए (n=7) है। परीक्षा में घटती घातों को सावधानी से गिनें।

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Mathematics Answer, Explanation and Revision Hints

गुणोत्तर श्रेणी \(729,243,81,27,\ldots\) में (1) कौन-सा पद है? / In the geometric progression \(729,243,81,27,\ldots\), which term is (1)?

Correct Answer: A. सातवाँ पद / (7)th term. Explanation: (729\left\(\frac{1}{3}\right\)^{n-1}=1) से \(3^{6-(n-1)}=1\) इसलिए (n=7) है। परीक्षा में घटती घातों को सावधानी से गिनें। / From (729\left\(\frac{1}{3}\right\)^{n-1}=1), \(3^{6-(n-1)}=1\), so (n=7). In exams count decreasing powers carefully.

Which concept should I revise for this Mathematics MCQ?

From (729\left\(\frac{1}{3}\right\)^{n-1}=1), \(3^{6-(n-1)}=1\), so (n=7). In exams count decreasing powers carefully.

What exam hint can help solve this Mathematics question?

(729\left\(\frac{1}{3}\right\)^{n-1}=1) से \(3^{6-(n-1)}=1\) इसलिए (n=7) है। परीक्षा में घटती घातों को सावधानी से गिनें।