Question

If the 2nd,3rd and 4th terms in the expansion of ${(a+x)}^n$ are respectively $240,720,1080$, find $a,x,n$.

Answer 1

Given that ${{}^{n}C}_1{x}^{n-1}a=240$ , ${{}^{n}C}_2{x}^{n-2}{a}^2=720$ , ${{}^{n}C}_3{x}^{n-3}{a}^3=1080$

By dividing second equation by first equation , we get $(n-1)\frac{a}{x}=6$

By dividing third equation by second equation , we get $(n-2)\frac{a}{x}=\frac{9}{2}$

By dividing above two ones , we get $\frac{n-2}{n-1}=\frac{9}{12}=\frac{3}{4}$

Therefore $4n-8=3n-3$ , which implies $n=5$ and $\frac{a}{x}=\frac{6}{4}=\frac{3}{2}$

from first equation , we get ${{}^{5}C}_1{x}^{4}a=240$

$\Rightarrow x^{4}a=48$

we know that $\frac{a}{x}=\frac{3}{2}$ , which gives $a=\frac{3x}{2}$

By substituting value of $a$ in $x^{4}a=48$ , we get $x^5=32$ , which gives $x=2$ and $a=3$