The `4^(th)` and the `7^(th)` terms of a G.P. are `(1)/(27)` and `(1)/(729)` respectively. Find the sum of n terms of this G.P.

Answer

To solve the problem, we need to find the sum of the first n terms of a geometric progression (G.P.) given that the 4th term \( T_4 \) is \( \frac{1}{27} \) and the 7th term \( T_7 \) is \( \frac{1}{729} \).

### Step-by-Step Solution:

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

2. **Set Up Equations for Given Terms**:
For the 4th term:
\[ T_4 = a \cdot r^{4-1} = a \cdot r^3 = \frac{1}{27} \quad \text{(Equation 1)} \]
For the 7th term:
\[ T_7 = a \cdot r^{7-1} = a \cdot r^6 = \frac{1}{729} \quad \text{(Equation 2)} \]

3. **Divide Equation 2 by Equation 1**:
To eliminate \( a \), divide Equation 2 by Equation 1:
\[ \frac{T_7}{T_4} = \frac{a \cdot r^6}{a \cdot r^3} = \frac{r^6}{r^3} = r^3 \]
Substituting the values:
\[ \frac{\frac{1}{729}}{\frac{1}{27}} = r^3 \]
Simplifying the left side:
\[ \frac{1}{729} \cdot 27 = r^3 \implies \frac{27}{729} = r^3 \implies \frac{1}{27} = r^3 \]

4. **Solve for r**:
Taking the cube root:
\[ r = \frac{1}{3} \]

5. **Substitute r back to find a**:
Substitute \( r \) back into Equation 1 to find \( a \):
\[ a \cdot \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
This simplifies to:
\[ a \cdot \frac{1}{27} = \frac{1}{27} \]
Therefore:
\[ a = 1 \]

6. **Find the Sum of the First n Terms**:
The formula for the sum of the first n terms \( S_n \) of a G.P. is:
\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \]
Substituting \( a = 1 \) and \( r = \frac{1}{3} \):
\[ S_n = 1 \cdot \frac{\left(\frac{1}{3}\right)^n - 1}{\frac{1}{3} - 1} \]
Simplifying the denominator:
\[ S_n = \frac{\left(\frac{1}{3}\right)^n - 1}{\frac{1}{3} - 1} = \frac{\left(\frac{1}{3}\right)^n - 1}{-\frac{2}{3}} = -\frac{3}{2} \left(\left(\frac{1}{3}\right)^n - 1\right) \]
Rearranging gives:
\[ S_n = \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]

### Final Answer:
Thus, the sum of the first n terms of the G.P. is:
\[ S_n = \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]