The sum of n terms of a progression is `(2^(n)-1)`. Show that it is a GP and find its common ratio.

Explanation

To solve the problem, we need to show that the given sum of n terms, \( S_n = 2^n - 1 \), represents a geometric progression (GP) and find its common ratio.

### Step-by-Step Solution:

1. **Identify the Sum of n Terms**:
We are given the sum of n terms of a progression:
\[ S_n = 2^n - 1 \]

2. **Find the (n-1)th Term**:
To find the nth term \( T_n \), we first need to calculate \( S_{n-1} \):
\[ S_{n-1} = 2^{n-1} - 1 \]

3. **Calculate the nth Term**:
The nth term \( T_n \) can be calculated using the formula:
\[ T_n = S_n - S_{n-1} \]
Substituting the values we have:
\[ T_n = (2^n - 1) - (2^{n-1} - 1) \]
Simplifying this gives:
\[ T_n = 2^n - 1 - 2^{n-1} + 1 = 2^n - 2^{n-1} \]
\[ T_n = 2^{n-1}(2 - 1) = 2^{n-1} \]

4. **Find the First Few Terms**:
Now, let's calculate the first few terms of the sequence:
- For \( n = 1 \):
\[ T_1 = 2^{1-1} = 2^0 = 1 \]
- For \( n = 2 \):
\[ T_2 = 2^{2-1} = 2^1 = 2 \]
- For \( n = 3 \):
\[ T_3 = 2^{3-1} = 2^2 = 4 \]
- For \( n = 4 \):
\[ T_4 = 2^{4-1} = 2^3 = 8 \]

The first few terms are \( 1, 2, 4, 8 \).

5. **Check for Common Ratio**:
To check if this is a GP, we need to find the common ratio \( r \):
\[ r = \frac{T_2}{T_1} = \frac{2}{1} = 2 \]
\[ r = \frac{T_3}{T_2} = \frac{4}{2} = 2 \]
\[ r = \frac{T_4}{T_3} = \frac{8}{4} = 2 \]

Since the ratio is constant, the sequence is indeed a geometric progression.

6. **Conclusion**:
We have shown that the sequence formed by the nth terms \( T_n = 2^{n-1} \) is a geometric progression with the first term \( T_1 = 1 \) and the common ratio \( r = 2 \).

### Final Answer:
The progression is a geometric progression (GP) with a common ratio of \( 2 \).