If the 2nd, 3rd and 4th terms in the expansion of

In the expansion of $(x+a)^{n}$, the $2 \mathrm{nd}, 3$ rd and 4 th terms are ${ }^{n} \mathrm{C}_{1} x^{\mathrm{n}-1} a^{1},{ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{x}^{\mathrm{n}-2} \mathrm{a}^{2}$ and ${ }^{n} C_{3} x^{\mathrm{n}-3} a^{3}$, respectively.

According to the question,

${ }^{n} C_{1} x^{n-1} a^{1}=240$

${ }^{n} C_{2} x^{n-2} a^{2}=720$

${ }^{n} C_{3} x^{n-3} a^{3}=1080$

$\Rightarrow \frac{{ }^{n} C_{2} x^{n-2} a^{2}}{{ }^{n} C_{1} x^{n-1} a^{1}}=\frac{720}{240}$

$\Rightarrow \frac{n-1}{2 x} a=3$

$\Rightarrow \frac{a}{x}=\frac{6}{n-1} \quad \ldots(1)$

Also,

$\frac{{ }^{n} C_{3} x^{n-3} a^{3}}{{ }^{n} C_{2} x^{n-2} a^{2}}=\frac{1080}{720}$

$\Rightarrow \frac{n-2}{3 x} a=\frac{3}{2}$

$\Rightarrow \frac{a}{x}=\frac{9}{2 n-4} \quad \ldots(2)$

Using $(1)$ and $(2)$ we get

$\frac{6}{n-1}=\frac{9}{2 n-4}$

$\Rightarrow n=5$

Putting in eqn (1) we get

$\Rightarrow 2 a=3 x$

Now, ${ }^{5} C_{1} x^{5-1}\left(\frac{3}{2} x\right)=240$

$\Rightarrow 15 x^{5}=480$

$\Rightarrow x^{5}=32$

$\Rightarrow x=2$

By putting the value of $x$ and $n$ in (1) we get

$a=3$