`3^(rd)` term of a G.P is 27 and its `6^(th)` term is 729, find the product of its first and `7^(th)` terms.

Answer

To solve the problem, we need to find the product of the first and the seventh terms of a geometric progression (G.P.) given that the third term is 27 and the sixth term is 729.

### Step-by-step Solution:

1. **Identify the terms of the G.P.:**
The n-th term of a G.P. can be expressed as:
\[ T_n = ar^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Write the equations for the given terms:**
- For the 3rd term:
\[ T_3 = ar^{3-1} = ar^2 = 27 \quad \text{(1)} \]
- For the 6th term:
\[ T_6 = ar^{6-1} = ar^5 = 729 \quad \text{(2)} \]

3. **Divide equation (2) by equation (1):**
This helps to eliminate \( a \):
\[ \frac{T_6}{T_3} = \frac{ar^5}{ar^2} = \frac{729}{27} \]
Simplifying the left side:
\[ r^{5-2} = r^3 = \frac{729}{27} \]
Calculating the right side:
\[ \frac{729}{27} = 27 \quad \Rightarrow \quad r^3 = 27 \]
Therefore, taking the cube root:
\[ r = 3 \quad \text{(since \( 3^3 = 27 \))} \]

4. **Substitute \( r \) back into equation (1) to find \( a \):**
Using \( r = 3 \) in equation (1):
\[ ar^2 = 27 \quad \Rightarrow \quad a(3^2) = 27 \]
Simplifying:
\[ 9a = 27 \quad \Rightarrow \quad a = \frac{27}{9} = 3 \]

5. **Find the first and seventh terms:**
- The first term \( T_1 \):
\[ T_1 = a = 3 \]
- The seventh term \( T_7 \):
\[ T_7 = ar^{7-1} = ar^6 = 3 \cdot 3^6 \]
Calculating \( 3^6 \):
\[ 3^6 = 729 \quad \Rightarrow \quad T_7 = 3 \cdot 729 = 2187 \]

6. **Calculate the product of the first and seventh terms:**
\[ \text{Product} = T_1 \cdot T_7 = 3 \cdot 2187 = 6561 \]

### Final Answer:
The product of the first and seventh terms is \( 6561 \).