Question

The 25th term of 9, 3, 1, \( \frac{1}{3} \), \( \frac{1}{9} \), ... is:

• A. \( \frac{1}{3^{24}} \)
• B. \( \frac{1}{3^{25}} \)
• C. \( \frac{1}{3^{23}} \)
• D. \( \frac{1}{3^{22}} \)
• E. \( \frac{1}{3^{26}} \)

Hint:

For a geometric sequence, the \(n\)-th term is calculated using the formula \(T_n = a \cdot r^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The given sequence is a Geometric Progression (GP) because the ratio between consecutive terms is constant. We need to find the specific term using the general formula for a GP.
Step 2: Key Formula or Approach
The $n$-th term of a GP is given by: \[ a_n = a \cdot r^{n-1} \] Where: $a$ is the first term. $r$ is the common ratio. $n$ is the term number.
Step 3: Detailed Calculation
1. Identify the values: - First term ($a$) = 9 - Common ratio ($r$) = $3/9 = 1/3$ - Term number ($n$) = 25 2. Apply the formula: - $a_{25} = 9 \cdot (1/3)^{25-1}$ - $a_{25} = 3^2 \cdot (1/3)^{24}$ - $a_{25} = 3^2 \cdot \frac{1}{3^{24}}$ - $a_{25} = \frac{1}{3^{24-2}} = \frac{1}{3^{22}}$
Step 4: Final Answer
The 25th term is \( \frac{1}{3^{22}} \).