Let `S_n` denote the sum of the first n terms of an AP
`S_(2n)=3S_n` Then the ratio `S_(3n)/S_n` is equal to

Options

A. 4
B. 6
C. 8
D. 10

Answer

To solve the problem, we start with the given information and use the formula for the sum of the first n terms of an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of an AP:**
The sum of the first n terms of an AP is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

2. **Finding \( S_{2n} \):**
Using the formula for \( S_n \), we can find \( S_{2n} \):
\[ S_{2n} = \frac{2n}{2} \times (2a + (2n - 1)d) = n \times (2a + (2n - 1)d) \]

3. **Setting Up the Given Condition:**
According to the problem, we have:
\[ S_{2n} = 3S_n \]
Substituting the expressions for \( S_{2n} \) and \( S_n \):
\[ n \times (2a + (2n - 1)d) = 3 \left( \frac{n}{2} \times (2a + (n - 1)d) \right) \]

4. **Simplifying the Equation:**
Cancelling \( n \) from both sides (assuming \( n \neq 0 \)):
\[ 2a + (2n - 1)d = \frac{3}{2} (2a + (n - 1)d) \]
Multiplying through by 2 to eliminate the fraction:
\[ 4a + (4n - 2)d = 3(2a + (n - 1)d) \]
Expanding the right side:
\[ 4a + (4n - 2)d = 6a + 3(n - 1)d \]
Rearranging gives:
\[ 4a + (4n - 2)d = 6a + 3nd - 3d \]
\[ 0 = 2a + (3n - 4n + 1)d \]
\[ 0 = 2a - nd + d \]
This can be rearranged to find a relationship between \( a \) and \( d \):
\[ 2a = (n - 1)d \quad \text{(1)} \]

5. **Finding \( S_{3n} \):**
Now, we calculate \( S_{3n} \):
\[ S_{3n} = \frac{3n}{2} \times (2a + (3n - 1)d) \]

6. **Finding the Ratio \( \frac{S_{3n}}{S_n} \):**
We need to find \( \frac{S_{3n}}{S_n} \):
\[ \frac{S_{3n}}{S_n} = \frac{\frac{3n}{2} \times (2a + (3n - 1)d)}{\frac{n}{2} \times (2a + (n - 1)d)} \]
Simplifying this gives:
\[ \frac{S_{3n}}{S_n} = \frac{3 \times (2a + (3n - 1)d)}{2a + (n - 1)d} \]

7. **Substituting from Equation (1):**
From equation (1), we can substitute \( 2a \) with \( (n - 1)d \):
\[ \frac{S_{3n}}{S_n} = \frac{3 \times ((n - 1)d + (3n - 1)d)}{(n - 1)d + (n - 1)d} \]
This simplifies to:
\[ \frac{S_{3n}}{S_n} = \frac{3 \times (4n - 2)d}{2(n - 1)d} \]
Cancelling \( d \) (assuming \( d \neq 0 \)):
\[ \frac{S_{3n}}{S_n} = \frac{3(4n - 2)}{2(n - 1)} = \frac{12n - 6}{2n - 2} = 6 \]

### Final Answer:
Thus, the ratio \( \frac{S_{3n}}{S_n} \) is equal to **6**.