If the 3^rd, 4^th and 5^th terms of ${(x+a)}^n$ are 720, 1080 and 810 respectively then $(x,a,n)$ is 

$T_3=720={}^n{\text{C}}_2{x}^{n-2}{a}^2\cdots \cdots \cdots \left(1\right)$

$T_4={}^n{\text{C}}_3{x}^{n-3}{a}^3=1080\cdots \cdots \cdots \left(2\right)$

$T_5={}^n{\text{C}}_4{x}^{n-4}{a}^4\cdots \cdots \cdots \left(3\right)$

$\frac{\left(2\right)}{\left(1\right)}=\frac{{}^n{\text{C}}_3{x}^{n-3}{a}^3}{{}^n{\text{C}}_2{x}^{n-2}{a}^2}=\frac{1080}{720}$

$\frac{n-2}{3}\left(\frac{9}{x}\right)=\frac{3}{2}\cdots \cdots \cdots \left(4\right)$

$\frac{\left(3\right)}{\left(2\right)}=\frac{{}^n{\text{C}}_4{x}^{n-4}{a}^4}{{}^n{\text{C}}_3{x}^{n-3}{a}^3}=\frac{810}{1080}$

$\left(\frac{n-3}{4}\right)\left(\frac{x}{a}\right)=\frac{3}{4}\cdots \cdots \cdots \left(5\right)$

$\frac{\left(4\right)}{\left(5\right)}\Rightarrow \frac{n-2}{3}\times \frac{4}{n-3}=\frac{3}{2}\times \frac{4}{3}$

$2(n-2)=3(n-3)$

$2n-4=3n-9$

$n=5$

$\text{from }\left(4\right)\Rightarrow \frac{a}{x}=\frac{3}{2}\Rightarrow a=\frac{3x}{2}$

$\text{from }\left(1\right)\Rightarrow 10x^3{\left(\frac{3x}{2}\right)}^2=720$

$x^5=2^5$

$x=2\&a=3$

$\therefore (x,a,n)=(2,3,5)$