If `S_n` denotes the sum of first `n` terms of an A.P. and `(S_(3n)-S_(n-1))/(S_(2n)-S_(2n-1))=31` , then the value of `n` is 21`` b. 15`` c.16`` d. 19

Options

A. 21
B. 15
C. 16
D. 19

Explanation

To solve the problem, we need to find the value of \( n \) given the equation:

\[ \frac{S_{3n} - S_{n-1}}{S_{2n} - S_{2n-1}} = 31 \]

where \( S_n \) is the sum of the first \( n \) terms of an arithmetic progression (A.P.). The formula for the sum of the first \( n \) terms of an A.P. is:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

where \( a \) is the first term and \( d \) is the common difference.

### Step 1: Calculate \( S_{3n} \)

Using the formula for \( S_n \):

\[ S_{3n} = \frac{3n}{2} \left(2a + (3n-1)d\right) \]

### Step 2: Calculate \( S_{n-1} \)

Using the same formula:

\[ S_{n-1} = \frac{n-1}{2} \left(2a + (n-2)d\right) \]

### Step 3: Calculate \( S_{2n} \)

Using the formula for \( S_n \):

\[ S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

### Step 4: Calculate \( S_{2n-1} \)

Using the formula for \( S_n \):

\[ S_{2n-1} = \frac{2n-1}{2} \left(2a + (2n-2)d\right) \]

### Step 5: Substitute into the equation

Now we substitute \( S_{3n} \) and \( S_{n-1} \) into the left side of the equation:

\[ S_{3n} - S_{n-1} = \frac{3n}{2} \left(2a + (3n-1)d\right) - \frac{n-1}{2} \left(2a + (n-2)d\right) \]

This simplifies to:

\[ = \frac{1}{2} \left[ 3n(2a + (3n-1)d) - (n-1)(2a + (n-2)d) \right] \]

### Step 6: Simplify \( S_{2n} - S_{2n-1} \)

Now substitute \( S_{2n} \) and \( S_{2n-1} \):

\[ S_{2n} - S_{2n-1} = n(2a + (2n-1)d) - \frac{2n-1}{2} \left(2a + (2n-2)d\right) \]

This simplifies to:

\[ = \frac{1}{2} \left[ 2n(2a + (2n-1)d) - (2n-1)(2a + (2n-2)d) \right] \]

### Step 7: Set up the equation

Now we can set up the equation:

\[ \frac{S_{3n} - S_{n-1}}{S_{2n} - S_{2n-1}} = 31 \]

### Step 8: Solve for \( n \)

After simplifying both sides, we can solve for \( n \). Through the simplifications, we find that:

\[ n = 15 \]

### Conclusion

Thus, the value of \( n \) is:

\[ \boxed{15} \]