$\Rightarrow \frac{{ }^n C_{r-1}(2)^{r-1}}{{ }^{ n } C_r(2)^r}=\frac{2}{5}$

$\Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5}$

$\Rightarrow \frac{r}{n-r+1}=\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$

$\Rightarrow \frac{{ }^n C_r(2)^r}{{ }^n C_{r+1}(2)^{r+1}}=\frac{5}{8}$

$\Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4}$

$\Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \ldots$

From (1) and (2)

$\Rightarrow 4 n +4=5 n -4 \Rightarrow n =8$

$(1)$ $\Rightarrow r=4$

so, coefficient of middle term is

${ }^8 C_4 2^4=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$