In the arithmetic progression whose common difference is non-zero, the sum of first 3n terms is equal to the sum of the next n terms. Then the ratio of the sum of the first 2n terms to the next 2n terms is

Options

A. `(1)/(5)`
B. `(2)/(3)`
C. `(3)/(4)`
D. none of these

Answer

To solve the problem, we need to find the ratio of the sum of the first 2n terms to the sum of the next 2n terms in an arithmetic progression (AP) where the sum of the first 3n terms is equal to the sum of the next n terms.

### Step-by-Step Solution:

1. **Define the Terms of the AP:**
Let the first term of the AP be \( a \) and the common difference be \( d \). The terms can be expressed as:
- First term: \( a_1 = a \)
- Second term: \( a_2 = a + d \)
- Third term: \( a_3 = a + 2d \)
- ...
- \( a_{3n} = a + (3n - 1)d \)
- \( a_{4n} = a + (4n - 1)d \)

2. **Sum of the First 3n Terms:**
The sum of the first \( 3n \) terms \( S_{3n} \) can be calculated using the formula for the sum of an arithmetic series:
\[ S_{3n} = \frac{3n}{2} \times (a + a_{3n}) = \frac{3n}{2} \times \left(a + \left(a + (3n - 1)d\right)\right) = \frac{3n}{2} \times (2a + (3n - 1)d) \]

3. **Sum of the Next n Terms:**
The sum of the next \( n \) terms \( S_n \) (from \( a_{3n+1} \) to \( a_{4n} \)) is:
\[ S_n = \frac{n}{2} \times (a_{3n+1} + a_{4n}) = \frac{n}{2} \times \left((a + 3nd) + (a + (4n - 1)d)\right) = \frac{n}{2} \times (2a + (7n - 1)d) \]

4. **Set the Sums Equal:**
According to the problem, we have:
\[ S_{3n} = S_n \]
Thus,
\[ \frac{3n}{2} \times (2a + (3n - 1)d) = \frac{n}{2} \times (2a + (7n - 1)d) \]

5. **Cancel Common Factors:**
We can cancel \( \frac{n}{2} \) from both sides (since \( n \neq 0 \)):
\[ 3 \times (2a + (3n - 1)d) = 2a + (7n - 1)d \]

6. **Expand and Simplify:**
Expanding both sides gives:
\[ 6a + 9nd - 3d = 2a + 7nd - d \]
Rearranging terms leads to:
\[ 4a + 2nd - 2d = 0 \]
Thus,
\[ 2a + nd - d = 0 \quad \Rightarrow \quad 2a = d - nd \quad \Rightarrow \quad a = \frac{d(1 - n)}{2} \]

7. **Sum of the First 2n Terms:**
Now, calculate the sum of the first \( 2n \) terms \( S_{2n} \):
\[ S_{2n} = \frac{2n}{2} \times (a + a_{2n}) = n \times \left(a + \left(a + (2n - 1)d\right)\right) = n \times (2a + (2n - 1)d) \]

8. **Sum of the Next 2n Terms:**
The sum of the next \( 2n \) terms \( S_{2n}' \) (from \( a_{2n+1} \) to \( a_{4n} \)):
\[ S_{2n}' = n \times \left(a_{2n+1} + a_{4n}\right) = n \times \left((a + 2nd) + (a + (4n - 1)d)\right) = n \times (2a + (6n - 1)d) \]

9. **Calculate the Ratio:**
The ratio \( \frac{S_{2n}}{S_{2n}'} \) is:
\[ \frac{S_{2n}}{S_{2n}'} = \frac{n(2a + (2n - 1)d)}{n(2a + (6n - 1)d)} = \frac{2a + (2n - 1)d}{2a + (6n - 1)d} \]

10. **Substituting for \( a \):**
Substitute \( a = \frac{d(1 - n)}{2} \):
\[ \frac{2\left(\frac{d(1 - n)}{2}\right) + (2n - 1)d}{2\left(\frac{d(1 - n)}{2}\right) + (6n - 1)d} \]
Simplifying gives:
\[ \frac{d(1 - n) + (2n - 1)d}{d(1 - n) + (6n - 1)d} = \frac{(1 - n + 2n - 1)d}{(1 - n + 6n - 1)d} = \frac{(n)d}{(5n)d} = \frac{1}{5} \]

### Final Answer:
The ratio of the sum of the first \( 2n \) terms to the sum of the next \( 2n \) terms is \( \frac{1}{5} \).