The 5th and 8th terms of a geometric sequence of real numbers are 7! And 8! Respectively. If the sum to first `n` tems of the G.P. is 2205, then `n` equals_______.

Explanation

To solve the problem, we need to find the value of \( n \) given that the 5th and 8th terms of a geometric progression (GP) are \( 7! \) and \( 8! \) respectively, and the sum of the first \( n \) terms is 2205.

### Step-by-Step Solution:

1. **Identify the terms of the GP**:
- The \( n \)-th term of a GP is given by the formula:
\[ T_n = a r^{n-1} \]
- For the 5th term:
\[ T_5 = a r^{4} = 7! \]
- For the 8th term:
\[ T_8 = a r^{7} = 8! \]

2. **Set up the equations**:
- We have two equations:
\[ a r^{4} = 7! \quad (1) \]
\[ a r^{7} = 8! \quad (2) \]

3. **Divide the equations to eliminate \( a \)**:
- Dividing equation (2) by equation (1):
\[ \frac{a r^{7}}{a r^{4}} = \frac{8!}{7!} \]
- This simplifies to:
\[ r^{3} = \frac{8!}{7!} = 8 \]
- Therefore:
\[ r^{3} = 8 \implies r = 2 \]

4. **Substitute \( r \) back to find \( a \)**:
- Substitute \( r = 2 \) into equation (1):
\[ a (2^{4}) = 7! \]
- This becomes:
\[ a \cdot 16 = 5040 \quad (\text{since } 7! = 5040) \]
- Thus:
\[ a = \frac{5040}{16} = 315 \]

5. **Use the sum formula for the first \( n \) terms of the GP**:
- The sum \( S_n \) of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a (r^{n} - 1)}{r - 1} \]
- We know \( S_n = 2205 \), \( a = 315 \), and \( r = 2 \):
\[ 2205 = \frac{315 (2^{n} - 1)}{2 - 1} \]
- This simplifies to:
\[ 2205 = 315 (2^{n} - 1) \]
- Dividing both sides by 315:
\[ \frac{2205}{315} = 2^{n} - 1 \]
- Calculating \( \frac{2205}{315} \):
\[ 2205 \div 315 = 7 \]
- Thus:
\[ 7 = 2^{n} - 1 \]
- Therefore:
\[ 2^{n} = 8 \]

6. **Solve for \( n \)**:
- Since \( 2^{3} = 8 \):
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( \boxed{3} \).