Question

The 25th term of $9, 3, 1, \frac{1}{3}, \frac{1}{9}, \ldots$ 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:

Express the first term as a power of the common ratio to simplify calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given sequence is a Geometric Progression (G.P.) because each term is obtained by multiplying the previous term by a constant factor, known as the common ratio (r). We need to find the 25th term of this sequence.
Step 2: Key Formula or Approach:
The formula for the n-th term (\(a_n\)) of a G.P. is:
\[ a_n = a \cdot r^{(n-1)} \] where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Step 3: Detailed Explanation:
First, we identify the first term and the common ratio from the sequence \( 9, 3, 1, \frac{1}{3}, \dots \).
The first term is \( a = 9 \).
The common ratio \(r\) can be found by dividing any term by its preceding term:
\[ r = \frac{3}{9} = \frac{1}{3} \] We can verify this with the next pair of terms: \( r = \frac{1}{3} \).
We need to find the 25th term, so \( n = 25 \).
Now, we plug these values into the formula:
\[ a_{25} = a \cdot r^{(25-1)} = 9 \cdot \left(\frac{1}{3}\right)^{24} \] To simplify, we can write \( 9 \) as \( 3^2 \).
\[ a_{25} = 3^2 \cdot \frac{1}{3^{24}} \] Using the exponent rule \( \frac{x^m}{x^n} = x^{m-n} \):
\[ a_{25} = \frac{3^2}{3^{24}} = 3^{2-24} = 3^{-22} \] This can be written as:
\[ a_{25} = \frac{1}{3^{22}} \] Step 4: Final Answer:
The 25th term of the G.P. is \( \frac{1}{3^{22}} \).