If S is the sum to infinite terms of a G.P whose first term is 'a', then the sum of the first n terms is

Options

A. `S(1-(a)/(S))^(n)`
B. `S{1-(1-(a)/(S))^(n)}`
C. `a{1-(1-(a)/(S))^(n)}`
D. none of these

Explanation

To find the sum of the first \( n \) terms of a geometric progression (G.P.) whose first term is \( a \) and whose sum to infinite terms is \( S \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Common Ratio**:
Let \( r \) be the common ratio of the G.P. The sum of the infinite terms of a G.P. is given by the formula:
\[ S = \frac{a}{1 - r} \]
This formula is valid only when \( |r| < 1 \).

2. **Rearranging the Formula**:
From the formula for \( S \), we can express \( r \) in terms of \( a \) and \( S \):
\[ S(1 - r) = a \]
Rearranging gives:
\[ S - Sr = a \implies Sr = S - a \implies r = 1 - \frac{a}{S} \]

3. **Sum of the First \( n \) Terms**:
The sum of the first \( n \) terms \( S_n \) of a G.P. can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

4. **Substituting \( r \)**:
Now substitute \( r = 1 - \frac{a}{S} \) into the formula for \( S_n \):
\[ S_n = \frac{a(1 - (1 - \frac{a}{S})^n)}{1 - (1 - \frac{a}{S})} \]
The denominator simplifies to:
\[ 1 - (1 - \frac{a}{S}) = \frac{a}{S} \]
Therefore, we can rewrite \( S_n \):
\[ S_n = \frac{a(1 - (1 - \frac{a}{S})^n)}{\frac{a}{S}} = S(1 - (1 - \frac{a}{S})^n) \]

5. **Final Expression**:
Thus, the sum of the first \( n \) terms of the G.P. is:
\[ S_n = S \left(1 - \left(1 - \frac{a}{S}\right)^n\right) \]

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