If `S_(n)` is the sum of the first n terms of an A.P. then : (a) `S_(3n)=3(S_(2n)-S_n)` (b) `S_(3n)=S_n+S_(2n)` (c) `S_(3n)=2(S_(2n)-S_(n)` (d) none of these

Options

A. `S_(3n)=3(S_(2n)-S_(n))`
B. `2.S_(3n)=S_(n)+S_(2n)`
C. `S_(3n)=2(S_(2n)-S_(n))`
D. none of these

Answer

To solve the problem, we need to analyze the sum of the first \( n \) terms of an arithmetic progression (A.P.) and derive the required relationships. Let's denote the first term of the A.P. as \( a \) and the common difference as \( d \).

### Step-by-Step Solution:

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

2. **Sum of the first 2n terms**:
Using the same formula, the sum of the first \( 2n \) terms \( S_{2n} \) is:
\[ S_{2n} = \frac{2n}{2} \left(2a + (2n - 1)d\right) = n \left(2a + (2n - 1)d\right) \]

3. **Sum of the first 3n terms**:
Similarly, the sum of the first \( 3n \) terms \( S_{3n} \) is:
\[ S_{3n} = \frac{3n}{2} \left(2a + (3n - 1)d\right) \]

4. **Finding \( S_{3n} \) in terms of \( S_{2n} \) and \( S_n \)**:
We need to compare \( S_{3n} \) with \( S_{2n} \) and \( S_n \). We start by calculating \( S_{2n} - S_n \):
\[ S_{2n} - S_n = n \left(2a + (2n - 1)d\right) - \frac{n}{2} \left(2a + (n - 1)d\right) \]
Simplifying this:
\[ = n \left(2a + 2nd - d\right) - \frac{n}{2} \left(2a + nd - d\right) \]
\[ = n \left(2a + 2nd - d\right) - \frac{n}{2} \left(2a + nd - d\right) \]

5. **Multiplying by 3**:
Now, we multiply \( S_{2n} - S_n \) by 3:
\[ 3(S_{2n} - S_n) = 3n \left(2a + 2nd - d\right) - \frac{3n}{2} \left(2a + nd - d\right) \]

6. **Comparing with \( S_{3n} \)**:
We can now equate this to \( S_{3n} \):
\[ S_{3n} = \frac{3n}{2} \left(2a + (3n - 1)d\right) \]

7. **Final Verification**:
After simplification, we find that:
\[ S_{3n} = 3(S_{2n} - S_n) \]
Thus, the correct option is:
\[ \text{(a) } S_{3n} = 3(S_{2n} - S_n) \]