If `n!,3xx(n!)" and "(n+1)!` are in GP, then the value of n will be

Options

A. 3
B. 4
C. 8
D. 10

Explanation

To solve the problem, we need to determine the value of \( n \) for which \( n! \), \( 3n! \), and \( (n+1)! \) are in geometric progression (GP).

### Step-by-Step Solution:

1. **Understanding the condition for GP**:
For three terms \( A \), \( B \), and \( C \) to be in GP, the condition is:
\[ B^2 = A \cdot C \]
Here, let \( A = n! \), \( B = 3n! \), and \( C = (n+1)! \).

2. **Substituting the values**:
Substitute \( A \), \( B \), and \( C \) into the GP condition:
\[ (3n!)^2 = n! \cdot (n+1)! \]

3. **Expanding both sides**:
Expanding the left side:
\[ (3n!)^2 = 9(n!)^2 \]
For the right side, recall that \( (n+1)! = (n+1) \cdot n! \):
\[ n! \cdot (n+1)! = n! \cdot (n+1) \cdot n! = (n+1)(n!)^2 \]

4. **Setting the equations equal**:
Now we have:
\[ 9(n!)^2 = (n+1)(n!)^2 \]

5. **Dividing both sides by \( (n!)^2 \)**:
Since \( n! \) is not zero for \( n \geq 0 \), we can safely divide both sides by \( (n!)^2 \):
\[ 9 = n + 1 \]

6. **Solving for \( n \)**:
Rearranging gives:
\[ n = 9 - 1 = 8 \]

### Final Answer:
Thus, the value of \( n \) is \( 8 \).