Let `S_n` denote the sum of first `n` terms of an A.P. If `S_(2n)=3S_n ,` then find the ratio `S_(3n)//S_ndot`

Explanation

To solve the problem, we need to find the ratio \( \frac{S_{3n}}{S_n} \) given that \( S_{2n} = 3S_n \).

### Step-by-step Solution:

1. **Understanding the Sum of an A.P.:**
The sum of the first \( n \) terms of an arithmetic progression (A.P.) can be expressed using the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Expressing \( S_{2n} \) and \( S_n \):**
Using the formula for the sum of an A.P., we can write:
\[ S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]
and
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

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

4. **Simplifying the Equation:**
The equation simplifies to:
\[ n \left(2a + (2n-1)d\right) = \frac{3n}{2} \left(2a + (n-1)d\right) \]
Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ 2a + (2n-1)d = \frac{3}{2} \left(2a + (n-1)d\right) \]

5. **Clearing the Fraction:**
Multiply through by 2 to eliminate the fraction:
\[ 4a + (4n-2)d = 3(2a + (n-1)d) \]
Expanding the right side:
\[ 4a + (4n-2)d = 6a + 3(n-1)d \]

6. **Rearranging Terms:**
Rearranging gives:
\[ 4a + 4nd - 2d = 6a + 3nd - 3d \]
Simplifying:
\[ 4a - 6a + 4nd - 3nd + 3d - 2d = 0 \]
This leads to:
\[ -2a + nd + d = 0 \quad \Rightarrow \quad 2a = (n + 1)d \]

7. **Finding \( S_{3n} \):**
Now we need to find \( S_{3n} \):
\[ S_{3n} = \frac{3n}{2} \left(2a + (3n-1)d\right) \]
Substituting \( 2a = (n + 1)d \):
\[ S_{3n} = \frac{3n}{2} \left((n + 1)d + (3n - 1)d\right) = \frac{3n}{2} \left((n + 1 + 3n - 1)d\right) = \frac{3n}{2} \left(4nd\right) \]
Thus,
\[ S_{3n} = 6n^2d \]

8. **Finding the Ratio \( \frac{S_{3n}}{S_n} \):**
Now we can find the ratio:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) = \frac{n}{2} \left((n + 1)d + (n - 1)d\right) = \frac{n}{2} \left(2nd\right) = n^2d \]
Therefore,
\[ \frac{S_{3n}}{S_n} = \frac{6n^2d}{n^2d} = 6 \]

### Final Answer:
The ratio \( \frac{S_{3n}}{S_n} \) is \( 6 \).