Binomial Theorem

Let the coefficients be $(n+5) c_{r-1}, (n+5) c_{r}$ and $(n+5) c_{r+1}$ .

then:

$\frac{(n+5) c_{r-1}}{(n+5) c_{r}} = \frac{5}{10}$

$\frac{r}{n+6-r} = \frac{1}{2}$

$r=\frac{n+6}{3}$

Also,

$\frac{(n+5) c_{r}}{(n+5) c_{r+1}} = \frac{10}{14}$

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

$12r = 5n + 18$

substituting value of r, we get $\boxed{n = 6}$