यदि \(a_n=9\left(\frac{1}{3}\right)^{n-1}\) है, तो \(a_3\) क्या होगा?

If \(a_n=9\left(\frac{1}{3}\right)^{n-1}\), what is \(a_3\)?

Author: Muft Shiksha Editorial Team Published: Updated:
Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

\(a_3=9\left(\frac{1}{3}\right)^2=1\). Calculate powers of fractional ratios carefully.

Step 2

Why this answer is correct

The correct answer is A. (1). \(a_3=9\left(\frac{1}{3}\right)^2=1\). Calculate powers of fractional ratios carefully.

Step 3

Exam Tip

\(a_3=9\left(\frac{1}{3}\right)^2=1\)। भिन्न अनुपात में घात की गणना सावधानी से करें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(a_n=9\left(\frac{1}{3}\right)^{n-1}\) है, तो \(a_3\) क्या होगा? / If \(a_n=9\left(\frac{1}{3}\right)^{n-1}\), what is \(a_3\)?

Correct Answer: A. (1). Explanation: \(a_3=9\left(\frac{1}{3}\right)^2=1\)। भिन्न अनुपात में घात की गणना सावधानी से करें। / \(a_3=9\left(\frac{1}{3}\right)^2=1\). Calculate powers of fractional ratios carefully.

Which concept should I revise for this Mathematics MCQ?

\(a_3=9\left(\frac{1}{3}\right)^2=1\). Calculate powers of fractional ratios carefully.

What exam hint can help solve this Mathematics question?

\(a_3=9\left(\frac{1}{3}\right)^2=1\)। भिन्न अनुपात में घात की गणना सावधानी से करें।