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

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Given that second term $T_{2}=240$

We have    ${T_2} = {\,^n}{C_1}{x^{n - 1}} \cdot a$

So       ${\,^n}{C_1}{x^{n - 1}} \cdot a = 240$        ..........$(1)$

Similarly    ${\,^n}{C_2}{x^{n - 2}}{a^2} = 720$         ...........$(2)$

and       $^{n} C_{x} x^{n-3} a^{3}=1080$             .............$(3)$

Dividing $(2)$ by $(1),$ we get

 $\frac{{{\,^n}{C_2}{x^{n - 2}}{a^2}}}{{^n{C_1}{x^{n - 1}}a}} = \frac{{720}}{{240}}$ i.e., $\frac{(n-1) !}{(n-2) !} \cdot \frac{a}{x}=6$

or      $\frac{a}{x}=\frac{6}{(n-1)}$           ...........$(4)$

Dividing $(3)$ by $(2),$ we have

$\frac{a}{x}=\frac{9}{2(n-2)}$         ...........$(5)$

From $(4)$ and $(5),$ 

$\frac{6}{n-1}=\frac{9}{2(n-2)}$        Thus, $n=5$

Hence, from $(1), 5 x^{4} a=240,$ and from $(4), \frac{a}{x}=\frac{3}{2}$

Solving these equations for $a$ and $x,$ we get $x=2$ and $a=3$

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