If p is a real number and if the middle term in the expansion of `((p)/(2)+2)^(8)` is 1120, find p.

Answer

To solve the problem, we need to find the value of \( p \) such that the middle term in the expansion of \( \left( \frac{p}{2} + 2 \right)^8 \) is equal to 1120.

### Step-by-Step Solution:

1. **Identify the Terms**:
We have the expression \( \left( \frac{p}{2} + 2 \right)^8 \). Here, we can identify:
- \( A = \frac{p}{2} \)
- \( B = 2 \)
- \( N = 8 \)

2. **Determine the Middle Term**:
Since \( N = 8 \) (which is even), the middle term can be found using the formula for the middle term in a binomial expansion:
\[ \text{Middle Term} = T_{r+1} = T_{\frac{N}{2} + 1} = T_{4 + 1} = T_5 \]

3. **Use the General Term Formula**:
The general term in the expansion of \( (A + B)^N \) is given by:
\[ T_{r+1} = \binom{N}{r} A^{N-r} B^r \]
For \( T_5 \) (where \( r = 4 \)):
\[ T_5 = \binom{8}{4} A^{8-4} B^4 = \binom{8}{4} A^4 B^4 \]

4. **Substituting Values**:
Substitute \( A \) and \( B \):
\[ T_5 = \binom{8}{4} \left( \frac{p}{2} \right)^4 (2)^4 \]

5. **Calculate \( \binom{8}{4} \)**:
\[ \binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

6. **Calculate \( T_5 \)**:
Now substitute \( \binom{8}{4} \) into the equation:
\[ T_5 = 70 \left( \frac{p}{2} \right)^4 (2^4) = 70 \left( \frac{p}{2} \right)^4 (16) \]
Simplifying gives:
\[ T_5 = 70 \times 16 \times \left( \frac{p^4}{16} \right) = 70 p^4 \]

7. **Set Equal to 1120**:
We know from the problem statement that this middle term equals 1120:
\[ 70 p^4 = 1120 \]

8. **Solve for \( p^4 \)**:
Divide both sides by 70:
\[ p^4 = \frac{1120}{70} = 16 \]

9. **Find \( p \)**:
To find \( p \), take the fourth root of both sides:
\[ p = \pm 2 \]

### Final Answer:
Thus, the value of \( p \) is \( 2 \) or \( -2 \).