Let `S_(n)` be the sum of the first n terms of an arithmetic progression .If `S_(3n)=3S_(2n)` , then the value of `(S_(4n))/(S_(2n))`is :

Options

A. 6
B. 4
C. 2
D. 8

Answer

To solve the problem, we need to find the value of \(\frac{S_{4n}}{S_{2n}}\) given that \(S_{3n} = 3S_{2n}\).

### Step-by-Step Solution:

1. **Formula for the Sum of the First n Terms of an AP**:
The sum of the first \(n\) terms of an arithmetic progression (AP) is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \(a\) is the first term and \(d\) is the common difference.

2. **Express \(S_{3n}\) and \(S_{2n}\)**:
Using the formula, we can express \(S_{3n}\) and \(S_{2n}\):
\[ S_{3n} = \frac{3n}{2} \left(2a + (3n-1)d\right) \]
\[ S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

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

4. **Simplify the Equation**:
Dividing both sides by \(n\) (assuming \(n \neq 0\)):
\[ \frac{3}{2} \left(2a + (3n-1)d\right) = 3 \left(2a + (2n-1)d\right) \]
Now, multiply both sides by 2 to eliminate the fraction:
\[ 3 \left(2a + (3n-1)d\right) = 6 \left(2a + (2n-1)d\right) \]

5. **Expand Both Sides**:
Expanding both sides gives:
\[ 6a + 9nd - 3d = 12a + 6nd - 6d \]

6. **Rearranging Terms**:
Rearranging the equation:
\[ 6a + 9nd - 3d - 12a - 6nd + 6d = 0 \]
Simplifying further:
\[ -6a + 3nd + 3d = 0 \]
Dividing through by 3:
\[ -2a + nd + d = 0 \]
Thus, we find:
\[ 2a = nd + d \quad \Rightarrow \quad 2a = d(n + 1) \]

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

8. **Substituting \(2a\)**:
Substitute \(2a = d(n + 1)\) into \(S_{4n}\):
\[ S_{4n} = 2n \left(d(n + 1) + (4n-1)d\right) = 2n \left(d(n + 1 + 4n - 1)\right) = 2n \left(d(5n)\right) = 10nd^2 \]

9. **Finding \(\frac{S_{4n}}{S_{2n}}\)**:
Now, we calculate \(S_{2n}\):
\[ S_{2n} = n \left(2a + (2n-1)d\right) = n \left(d(n + 1) + (2n-1)d\right) = n \left(d(n + 1 + 2n - 1)\right) = n(3nd) = 3nd^2 \]
Therefore, we can find the ratio:
\[ \frac{S_{4n}}{S_{2n}} = \frac{10nd^2}{3nd^2} = \frac{10}{3} \]

### Final Answer:
\[ \frac{S_{4n}}{S_{2n}} = \frac{10}{3} \]