Question:
The 25th term of $9, 3, 1, \frac{1}{3}, \frac{1}{9}, \ldots$ is:
The 25th term of $9, 3, 1, \frac{1}{3}, \frac{1}{9}, \ldots$ is:
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Express the first term as a power of the common ratio to simplify calculations.
Updated On: Apr 28, 2026
- $\frac{1}{3^{24}}$
- $\frac{1}{3^{25}}$
- $\frac{1}{3^{23}}$
- $\frac{1}{3^{22}}$
- $\frac{1}{3^{26}}$
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The Correct Option is D
Solution and Explanation
Step 1: Analysis
The sequence is a G.P. with first term $a = 9$ and common ratio $r = \frac{3}{9} = \frac{1}{3}$.
Step 2: Formula
The $n$-th term is $a_n = ar^{n-1}$.
Step 3: Calculation
$a_{25} = 9 \times (\frac{1}{3})^{25-1} = 3^2 \times (\frac{1}{3})^{24} = \frac{3^2}{3^{24}} = \frac{1}{3^{22}}$. Final Answer: (D)
The sequence is a G.P. with first term $a = 9$ and common ratio $r = \frac{3}{9} = \frac{1}{3}$.
Step 2: Formula
The $n$-th term is $a_n = ar^{n-1}$.
Step 3: Calculation
$a_{25} = 9 \times (\frac{1}{3})^{25-1} = 3^2 \times (\frac{1}{3})^{24} = \frac{3^2}{3^{24}} = \frac{1}{3^{22}}$. Final Answer: (D)
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