यदि \(a_n=81\left(\frac{1}{3}\right)^{n-1}\) है, तो \(a_2+a_5\) कितना होगा?

If \(a_n=81\left(\frac{1}{3}\right)^{n-1}\), what is \(a_2+a_5\)?

Author: Muft Shiksha Editorial Team Published: Updated:
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Correct Answer

A. (28)

Step 1

Concept

\(a_2=27\) and \(a_5=1\), so the sum is (28). With a fractional ratio, terms decrease quickly.

Step 2

Why this answer is correct

The correct answer is A. (28). \(a_2=27\) and \(a_5=1\), so the sum is (28). With a fractional ratio, terms decrease quickly.

Step 3

Exam Tip

\(a_2=27\) और \(a_5=1\), इसलिए योग (28) है। भिन्न अनुपात में हर पद तेजी से घटता है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(a_n=81\left(\frac{1}{3}\right)^{n-1}\) है, तो \(a_2+a_5\) कितना होगा? / If \(a_n=81\left(\frac{1}{3}\right)^{n-1}\), what is \(a_2+a_5\)?

Correct Answer: A. (28). Explanation: \(a_2=27\) और \(a_5=1\), इसलिए योग (28) है। भिन्न अनुपात में हर पद तेजी से घटता है। / \(a_2=27\) and \(a_5=1\), so the sum is (28). With a fractional ratio, terms decrease quickly.

Which concept should I revise for this Mathematics MCQ?

\(a_2=27\) and \(a_5=1\), so the sum is (28). With a fractional ratio, terms decrease quickly.

What exam hint can help solve this Mathematics question?

\(a_2=27\) और \(a_5=1\), इसलिए योग (28) है। भिन्न अनुपात में हर पद तेजी से घटता है।