The coefficients of three consecutive terms of $(1+x{)}^{n+5}$ are in the ratio $5: 10: 14$. Then, $n =$

Let , $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ be three consecutive terms in the expansion of $(1+x{)}^{n+5}$

It is given that 

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

$\Rightarrow \frac{{ }^{n+5}{C}_r}{{ }^{1+3}{C}_{r-1}}=\frac{10}{5}$ and $\frac{{ }^{n+5}{C}_{r+1}}{{ }^{n+5}{C}_r}=\frac{14}{10}$

$\Rightarrow \frac{n+5-r+1}{r}=2$ and $\frac{n+5-r}{r+1}=\frac{7}{5}$ 

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

$\Rightarrow n=6,r=4$