Let the coefficients of three consecutive terms in the binomial expansion of ( 1 + 2 x ) n be in the ratio 2…

Given,

The coefficients of three consecutive terms in the binomial expansion of $(1+2x{)}^{\mathrm{n}}$ be in the ratio $2: 5: 8$.

Now $r^{th}$ term in expansion of $(1+2x{)}^{\mathrm{n}}$ is given by, ${\mathrm{t}}_{\mathrm{r}+1}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}(2\mathrm{x}{)}^{\mathrm{r}}$

Now let $T_r, T_{r+1}, T_{r+2}$ are in the ratio $2:5:8$

$\Rightarrow \frac{T_r}{T_{r+1}}=\frac{{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}-1}(2{)}^{\mathrm{r}-1}}{{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}(2{)}^{\mathrm{r}}}=\frac{2}{5}$

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

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

$\Rightarrow 9r=4(n+1) .......\left(1\right)$

Now taking other ratio $\frac{T_{r+1}}{T_{r+2}}=\frac{{}^{n}C_r(2{)}^{\mathrm{r}}}{{}^{n}C_{r+1}(2{)}^{\mathrm{r}+1}}=\frac{5}{8}$

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

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

$\Rightarrow 5n-4=9r .........\left(2\right)$

From equation $\left(1\right) \& \left(2\right)$ we get,

$4\mathrm{n}+4=5\mathrm{n}-4\Rightarrow \mathrm{n}=8$ and $r=4$

So, coefficient of middle term will be

${}^8\mathrm{C}_4{2}^4=16\times \frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=16\times 70=1120$