Let `S_(n)` denotes the sum of first n terms of an AP and `3S_(n)=S_(2n)`.
Q. What is `S_(3n):S_(n)` equal to?

Options

A. `4:1`
B. `6:1`
C. `8:1`
D. `10:1`

Answer

To solve the problem, we need to find the ratio \( S(3n) : S(n) \) given that \( 3S(n) = S(2n) \).

### Step-by-Step Solution:

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

2. **Expressing \( S(2n) \)**:
Using the formula for the sum of the first \( 2n \) terms:
\[ S(2n) = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

3. **Using the Given Relation**:
According to the problem, we have:
\[ 3S(n) = S(2n) \]
Substituting the expressions for \( S(n) \) and \( S(2n) \):
\[ 3 \left(\frac{n}{2} \left(2a + (n-1)d\right)\right) = n \left(2a + (2n-1)d\right) \]
Simplifying this gives:
\[ \frac{3n}{2} \left(2a + (n-1)d\right) = n \left(2a + (2n-1)d\right) \]

4. **Cancelling \( n \) (assuming \( n \neq 0 \))**:
Dividing both sides by \( n \):
\[ \frac{3}{2} \left(2a + (n-1)d\right) = 2a + (2n-1)d \]

5. **Expanding Both Sides**:
Expanding the left side:
\[ 3a + \frac{3}{2}(n-1)d = 2a + (2n-1)d \]
Rearranging gives:
\[ 3a - 2a + \frac{3}{2}(n-1)d - (2n-1)d = 0 \]
Simplifying further:
\[ a + \left(\frac{3}{2}(n-1) - (2n-1)\right)d = 0 \]
\[ a + \left(\frac{3n - 3 - 4n + 2}{2}\right)d = 0 \]
\[ a + \left(\frac{-n - 1}{2}\right)d = 0 \]
This leads to:
\[ a = \frac{(n + 1)}{2}d \]

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

7. **Finding the Ratio \( S(3n) : S(n) \)**:
Now we can find the ratio:
\[ S(n) = \frac{n}{2} \left(2a + (n-1)d\right) = \frac{n}{2} \left(n d + d\right) = \frac{n^2d + nd}{2} \]
Thus:
\[ S(n) = \frac{(n^2 + n)d}{2} \]
Now, the ratio:
\[ \frac{S(3n)}{S(n)} = \frac{6n^2d}{\frac{(n^2 + n)d}{2}} = \frac{6n^2d \cdot 2}{(n^2 + n)d} = \frac{12n^2}{n^2 + n} \]
Simplifying gives:
\[ = \frac{12n^2}{n(n + 1)} = \frac{12n}{n + 1} \]

### Final Answer:
The ratio \( S(3n) : S(n) \) is \( \frac{12n}{n + 1} \).