Question

The $2^{nd},3^{rd}$ and $4^{th}$ in the B.E ${\left(x+a\right)}^n$ are 240, 720 and 1080, respecttively . find x , a and n.

Answer 1

$(x+y{)}^n=x^n+\underset{240}{{}^n{C}_1{x}^{n-1}y}+\underset{720}{{}^n{C}_2{x}^{n-2}{y}^2}+\underset{1080}{{}^n{C}_3{x}^{n-3}{y}^3}+$.....

$n{x}^{n-1}y=240$ …….$\left(1\right)$

$\frac{n(n-1)}{2}{x}^{n-2}{y}^2=720$ ………$\left(2\right)$

$\frac{h(n-1)(n-2)}{6}{x}^{n-3}{y}^3=1080$ ……$\left(3\right)$

$\left(1\right)\times \left(3\right)$

$\Rightarrow \frac{n^2(n-1)(n-2)}{6}{x}^{2n-4}{y}^4=240\times 1080$ …….$\left(4\right)$

$\left\{\right(2\left)\right\}$

$\Rightarrow \frac{n^2(n-1{)}^2}{4}{x}^{2n-4}{y}^4=(720{)}^2$ ……$\left(5\right)$

Divide $\frac{4}{5}$

$\Rightarrow \frac{4(n-2)}{6(n-1)}=\frac{240\times 1080}{720\times 720}=\frac{1}{2}$

$\Rightarrow 4n-8=\frac{1}{2}(6n-6)$

$\Rightarrow 8n-16=6n-6\Rightarrow 2n=10$

$n=5$ Ans

Equation $\left(1\right)$ $n{x}^{n-1}y\Rightarrow 5x^{4}y=240$ $\Rightarrow x^{4}y=48$ …..$\left(6\right)$

Equation $\left(2\right)$ $\frac{n(n-1)}{2}{x}^{n-2}{y}^2=720\Rightarrow \frac{5\times 4}{2}{x}^3{y}^2=720$

$\Rightarrow x^3{y}^2=72$ ……$\left(7\right)$

Now $\frac{equation\left(6\right)}{equation\left(7\right)}\frac{(x^{4}y{)}^2}{x^3{y}^2}$

$\Rightarrow \frac{x^8{y}^2}{x^3{y}^2}=\frac{48\times 48}{72}=32$

$\Rightarrow x^5=32$

$\Rightarrow x=2$Answer

Now $x^{4}y=4p\Rightarrow 2^{4}xy=48$

$\Rightarrow y=3$

$\therefore n=5,x=2,y=3$.