If p is a real number and if the middle

Given expansion is $\left(\frac{p}{2}+2\right)^{8}$

Since index is $n=8$, there is only one middle term, i.e., $\left(\frac{8}{2}+1\right)$ th $=5^{\text {th }}$ term

$T_{5}=T_{4+1}={ }^{8} C_{4}\left(\frac{p}{2}\right)^{8-4} \cdot 2^{4}$

$\Rightarrow \quad 1120={ }^{8} C_{4} p^{4}$

By substituting the values, we get

$\Rightarrow \quad 1120=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} p^{4}$

$\Rightarrow 1120=7 \times 2 \times 5 \times p^{4}$

$\Rightarrow p^{4}=\frac{1120}{70}$

$\Rightarrow p^{4}=16$

$\Rightarrow p^{2}=4 \Rightarrow p=\pm 2$