The second,third and fourth terms in the binomial expansion $(x+a)^n$ are $240, 720$ and $1080$ respectively. Find $x, a$ and $n$.

(N/A) Given that the second term $T_2 = 240$.
We have $T_2 = ^nC_1 x^{n-1} a = 240$ ..........$(1)$
Similarly,$T_3 = ^nC_2 x^{n-2} a^2 = 720$ ..........$(2)$
And $T_4 = ^nC_3 x^{n-3} a^3 = 1080$ ..........$(3)$
Dividing $(2)$ by $(1)$:
$\frac{^nC_2 x^{n-2} a^2}{^nC_1 x^{n-1} a} = \frac{720}{240} \implies \frac{n-1}{2} \cdot \frac{a}{x} = 3 \implies \frac{a}{x} = \frac{6}{n-1}$ ..........$(4)$
Dividing $(3)$ by $(2)$:
$\frac{^nC_3 x^{n-3} a^3}{^nC_2 x^{n-2} a^2} = \frac{1080}{720} \implies \frac{n-2}{3} \cdot \frac{a}{x} = \frac{3}{2} \implies \frac{a}{x} = \frac{9}{2(n-2)}$ ..........$(5)$
Equating $(4)$ and $(5)$:
$\frac{6}{n-1} = \frac{9}{2(n-2)} \implies 12(n-2) = 9(n-1) \implies 12n - 24 = 9n - 9 \implies 3n = 15 \implies n = 5$.
Substituting $n=5$ into $(4)$:
$\frac{a}{x} = \frac{6}{5-1} = \frac{6}{4} = \frac{3}{2} \implies a = \frac{3x}{2}$.
Substituting $n=5$ and $a = \frac{3x}{2}$ into $(1)$:
$5x^4 \cdot (\frac{3x}{2}) = 240 \implies x^5 = 240 \cdot \frac{2}{15} = 16 \cdot 2 = 32 \implies x = 2$.
Then $a = \frac{3(2)}{2} = 3$.
Thus,$x=2, a=3, n=5$.