Consider an A.P. with first term 'a'. Let `S_(n)` denote the sum its terms. If `(S_(kx))/(S_(x))` is independent of x, then `S_(n)=`

Options

A. `n^(2)a`
B. na
C. `2n^(2)a`
D. `(n^(2)+n)a`

Explanation

To solve the problem, we need to find the sum of the first \( n \) terms of an arithmetic progression (A.P.) given that the ratio \( \frac{S_{kx}}{S_x} \) is independent of \( x \).

### Step-by-Step Solution:

1. **Define the first term and common difference**:
Let the first term of the A.P. be \( a \) and the common difference be \( d \).

2. **Write the formula for the sum of the first \( n \) terms**:
The sum of the first \( n \) terms \( S_n \) of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

3. **Express \( S_{kx} \) and \( S_x \)**:
Using the formula for \( S_n \):
\[ S_{kx} = \frac{kx}{2} \left(2a + (kx - 1)d\right) \]
\[ S_x = \frac{x}{2} \left(2a + (x - 1)d\right) \]

4. **Set up the ratio**:
Now we compute the ratio \( \frac{S_{kx}}{S_x} \):
\[ \frac{S_{kx}}{S_x} = \frac{\frac{kx}{2} \left(2a + (kx - 1)d\right)}{\frac{x}{2} \left(2a + (x - 1)d\right)} \]
Simplifying this gives:
\[ = \frac{k \left(2a + (kx - 1)d\right)}{2a + (x - 1)d} \]

5. **Analyze the independence of \( x \)**:
For the ratio \( \frac{S_{kx}}{S_x} \) to be independent of \( x \), the terms involving \( x \) in the numerator and denominator must cancel out. This can happen if the coefficient of \( x \) in both the numerator and denominator are equal.

6. **Set the coefficients equal**:
The coefficient of \( x \) in the numerator is \( kd \) and in the denominator is \( d \). For independence from \( x \), we set:
\[ kd = d \implies k = 1 \quad \text{(if \( d \neq 0 \))} \]

7. **Find the condition for \( d \)**:
We also need to ensure that the constant terms are equal:
\[ 2a - d = 0 \implies d = 2a \]

8. **Substitute \( d \) back into the sum formula**:
Now substituting \( d = 2a \) into the sum formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)(2a)\right) \]
Simplifying this gives:
\[ S_n = \frac{n}{2} \left(2a + 2an - 2a\right) = \frac{n}{2} \cdot 2an = n^2 a \]

9. **Final Result**:
Therefore, the sum of the first \( n \) terms \( S_n \) is:
\[ S_n = n^2 a \]