Let the coefficients of the three consecutive terms in the binomial expansion of ${\left(1+2\mathrm{x}\right)}^{\mathrm{n}}$ be in the ratio 2:5:8. Then the coefficient of the term, which is in the middle of these three terms, is ______.

$$\begin{gathered} \begin{array}{l}\text{ Coef }{\mathrm{T}}_{\mathrm{r}}={\mathrm{nc}}_{\mathrm{r}-1}{2}^{\mathrm{r}-1}\\ \text{ Coef }{\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}{2}^{\mathrm{r}}\\ \mathrm{Coef}{\mathrm{T}}_{\mathrm{r}+2}{=}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}+1}{2}^{\mathrm{r}+1}\\ \frac{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}\cdot 2^{\mathrm{r}}}{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}-1}\cdot 2^{\mathrm{r}-1}}=\frac{5}{2}\\ \frac{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}+1}\cdot 2^{\mathrm{r}+1}}{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}\cdot 2^{\mathrm{r}}}=\frac{8}{5}\\ \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{5}{4}\dots \dots ..(1) \\ \begin{array}{l}\frac{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}+1}\cdot 2^{\mathrm{r}+1}}{{ }^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}\cdot 2^{\mathrm{r}}}=\frac{8}{5}\\ \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{4}{5}\dots \dots .\left(2\right)\end{array}\\ \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{4}{5}\dots \dots .\\ 1\mathrm{\&}2\Rightarrow \frac{5\mathrm{r}}{4}-1=\frac{4(\mathrm{r}+1)}{5}\\ \Rightarrow 25\mathrm{r}-20=16\mathrm{r}+16\\ \Rightarrow 9\mathrm{r}=36\Rightarrow \mathrm{r}=4\\ \mathrm{n}=8\\ \text{ Required coeff }=8{\mathrm{c}}_4\times 2^4\\ =1120\\ \end{array} \end{gathered}$$