If `S_(n)` denotes the sum of first n terms of an A.P., then
`(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))` is equal to

Options

A. 2n-1
B. 2n+1
C. 4n+1
D. 2n+3

Explanation

To solve the problem, we need to find the value of the expression \((S_{3n} - S_{n-1}) / (S_{2n} - S_{n-1})\), where \(S_n\) denotes the sum of the first \(n\) terms of an arithmetic progression (A.P.).

### Step 1: Write the formula for the sum of the first \(n\) terms of an A.P.
The sum of the first \(n\) terms of an A.P. 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.

### Step 2: Calculate \(S_{3n}\)
Using the formula:
\[ S_{3n} = \frac{3n}{2} \left(2a + (3n-1)d\right) \]

### Step 3: Calculate \(S_{2n}\)
Using the formula:
\[ S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

### Step 4: Calculate \(S_{n-1}\)
Using the formula:
\[ S_{n-1} = \frac{n-1}{2} \left(2a + (n-2)d\right) \]

### Step 5: Substitute \(S_{3n}\), \(S_{2n}\), and \(S_{n-1}\) into the expression
Now substitute these sums into the expression:
\[ \frac{S_{3n} - S_{n-1}}{S_{2n} - S_{n-1}} \]

### Step 6: Calculate \(S_{3n} - S_{n-1}\)
\[ S_{3n} - S_{n-1} = \frac{3n}{2} \left(2a + (3n-1)d\right) - \frac{n-1}{2} \left(2a + (n-2)d\right) \]
Simplifying this gives:
\[ = \frac{1}{2} \left[ 3n(2a + (3n-1)d) - (n-1)(2a + (n-2)d) \right] \]

### Step 7: Calculate \(S_{2n} - S_{n-1}\)
\[ S_{2n} - S_{n-1} = n \left(2a + (2n-1)d\right) - \frac{n-1}{2} \left(2a + (n-2)d\right) \]
Simplifying this gives:
\[ = \frac{1}{2} \left[ 2n(2a + (2n-1)d) - (n-1)(2a + (n-2)d) \right] \]

### Step 8: Form the final expression
Now we can form the final expression:
\[ \frac{S_{3n} - S_{n-1}}{S_{2n} - S_{n-1}} = \frac{3n(2a + (3n-1)d) - (n-1)(2a + (n-2)d)}{2n(2a + (2n-1)d) - (n-1)(2a + (n-2)d)} \]

### Step 9: Simplify the expression
After simplification, we can find that the expression reduces to:
\[ \frac{3n}{2n - 1} \]

### Final Answer
Thus, the value of \(\frac{S_{3n} - S_{n-1}}{S_{2n} - S_{n-1}}\) is:
\[ \frac{3n}{2n - 1} \]