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

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

$\Rightarrow \frac{\left(n+5\right)!}{\left(r-1\right)!\left(n-r+6\right)!}:\frac{\left(n+5\right)!}{\left(r\right)!\left(n-r+5\right)!}:\frac{\left(n+5\right)!}{\left(r+1\right)!\left(n-r+4\right)!}=\text{5 : 10 :14}$

$\Rightarrow \frac{1}{\left(r-1\right)!\left(n-r+6\right)!}:\frac{1}{\left(r\right)!\left(n-r+5\right)!}:\frac{1}{\left(r+1\right)!\left(n-r+4\right)!}=\text{5 : 10 :14}$

$\Rightarrow \frac{\left(r+1\right)r}{\left(n-r+6\right)!}:\frac{r+1}{\left(n-r+5\right)!}:\frac{1}{\left(n-r+4\right)!}=\text{5 : 10 :14}$

$\Rightarrow \left(r+1\right)r : \left(r+1\right)\left(n-r+6\right) : \left(n-r+6\right)\left(n-r+5\right)=\text{5 : 10 :14}$

$\Rightarrow \frac{\left(r+1\right)\left(n-r+6\right)}{\left(r+1\right)r}=\frac{10}{5}\text{and}\frac{\left(n-r+6\right)\left(n-r+5\right)}{\left(r+1\right)\left(n-r+6\right)}=\frac{14}{10}$

$\Rightarrow \frac{\left(n-r+6\right)}{r}=2\text{and}\frac{\left(n-r+5\right)}{\left(r+1\right)}=\frac{7}{5}$

$\Rightarrow n=3r-6\text{and}\text{n}=\frac{12r-18}{5}$

$\Rightarrow 3r-6=\frac{12r-18}{5}=n$

$\Rightarrow 15r-30=12r-18$

$\Rightarrow r=4$

$\Rightarrow n=6$