The first term of a G.P. is 9 and `5^(th)` tems is 1/9. Which term of the G.P. is 1/729 ?

Options

A. 9
B. 8
C. 7
D. None of these

Answer

To solve the problem, we need to find out which term of the geometric progression (G.P.) is equal to \( \frac{1}{729} \), given that the first term \( a = 9 \) and the fifth term is \( \frac{1}{9} \).

### Step-by-Step Solution:

1. **Identify the first term and the fifth term:**
- The first term \( a = 9 \).
- The fifth term \( a_5 = \frac{1}{9} \).

2. **Use the formula for the \( n \)-th term of a G.P.:**
- The formula for the \( n \)-th term of a G.P. is given by:
\[ a_n = a \cdot r^{n-1} \]
- For the fifth term:
\[ a_5 = a \cdot r^{5-1} = a \cdot r^4 \]
- Substituting the known values:
\[ \frac{1}{9} = 9 \cdot r^4 \]

3. **Solve for \( r^4 \):**
- Rearranging the equation:
\[ r^4 = \frac{1}{9} \cdot \frac{1}{9} = \frac{1}{81} \]

4. **Find \( r \):**
- Taking the fourth root:
\[ r = \left(\frac{1}{81}\right)^{\frac{1}{4}} = \frac{1}{3} \]

5. **Now, we need to find which term is equal to \( \frac{1}{729} \):**
- Set up the equation:
\[ a_n = a \cdot r^{n-1} = 9 \cdot \left(\frac{1}{3}\right)^{n-1} \]
- We want to find \( n \) such that:
\[ 9 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{729} \]

6. **Simplify the equation:**
- Since \( 729 = 3^6 \), we can rewrite \( \frac{1}{729} \) as \( \frac{1}{3^6} \):
\[ 9 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{3^6} \]
- Rewrite \( 9 \) as \( 3^2 \):
\[ 3^2 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{3^6} \]
- This simplifies to:
\[ \frac{3^2}{3^{n-1}} = \frac{1}{3^6} \]
- Which can be rewritten as:
\[ 3^{2 - (n-1)} = 3^{-6} \]
- Therefore:
\[ 2 - (n-1) = -6 \]

7. **Solve for \( n \):**
- Rearranging gives:
\[ 2 + 6 = n - 1 \]
\[ n = 9 \]

### Conclusion:
The term of the G.P. that is equal to \( \frac{1}{729} \) is the **9th term**.