If for some positive integer $n$the coefficients of three consecutive  terms in the binomial  expansion of ${(1+x)}^{n+5}$are in the ratio $5: 10; 14,.$ then the largest coefficient- in the expansion is

Let three consecutive  terms be $T_r,T_{r+1}$ and $T_{r+2}$ 

Their coefficients are ${ }^{n+5}{C}_{r-1}{,}^{n+5}{C}_r$  and ${ }^{n+5}{C}_r+1$ .It is given that 

${ }^{n+5}{C}_{r-1}{:}^{n+5}{C}_r{:}^{n+5}{C}_{r+1}=5:10:14$

$\frac{{ }^{n+5}{C}_r}{{ }^{n+5}{C}_{r-1}}=\frac{10}{5}$ and $\frac{{ }^{n+5}{C}_{r+1}}{{ }^{n+5}{C}_r}=\frac{14}{10}$

$\Rightarrow \frac{n+5-r+1}{r}=2$ and $\frac{n+5-r}{r+1}=\frac{7}{5}\left[\because \frac{n_r}{n_{r-1}}=\frac{n-r+1}{r}\right]$

$\Rightarrow n-r+6=2r$ and $5(n-r+5)=7r+7$

$\Rightarrow n-3r+6=0$ and $5n-12r+18=0$

$\Rightarrow n=6,r=4$ 

Thus, the coefficients in the binomial expansion of $(1+x{)}^{11}$ are 

${ }^{11}{C}_0{,}^{11}{C}_1,\dots {,}^{11}{C}_{11}$.The greatest coefficient  is ${ }^{11}{C}_5{=}^{11}{C}_6=462$