Question

The third, fourth, and fifth terms in the expansion of $(x+a{)}^n$ in descending powers of x are 84, 280 and 560 respectively, find x, a and n.

• A. $x=-1,a=-2,n=2$
• B. $x=1,a=-2,n=-2$
• C. $x=-1,a=2,n=-2$
• D. $x=1,a=2,n=2$

Answer 1

The correct option is D $x=1,a=2,n=2$

By Binomial Expansion : For a expansion of $(x+a{)}^n$

$T_{r+1}{=}^n{C}_r{x}^{n-r}{a}^r$ $o\le r\le n$

Therefore , 3rd term : $T_3{=}^n{C}_2{x}^{n-2}{a}^2=84$.......(1)

Similarly, $T_4{=}^n{C}_3{x}^{x-3}{a}^3=280$........(2)

$T_5{=}^n{C}_4{x}^{x-4}{a}^4=560$.........(3)

$\frac{\left(2\right)}{\left(1\right)}\to \frac{280}{84}=\frac{{}^n{C}_3}{{}^n{C}_2}\left(\frac{a}{x}\right)\Rightarrow \frac{{}^n{C}_3}{{}^n{C}_2}\left(\frac{a}{x}\right)=\frac{10}{3}\to \left(4\right)$

$\frac{\left(3\right)}{\left(2\right)}\to \frac{560}{280}=\frac{{}^n{C}_4}{{}^n{C}_3}\left(\frac{a}{x}\right)\Rightarrow \frac{{}^n{C}_4}{{}^n{C}_3}\left(\frac{a}{x}\right)=2\to \left(5\right)$

$\left(4\right)\to \frac{n!}{(n-3)!3!}\times \frac{(n-2)!2!}{n!}\times \left(\frac{a}{x}\right)=\frac{a}{x}=\frac{10}{3}\Rightarrow \frac{n-2}{3}\left(\frac{a}{x}\right)=\frac{10}{3}$

$\left(5\right)\to \frac{n!}{(n-4)!4!}\times \frac{(n-3)!3!}{n!}\times \left(\frac{a}{x}\right)=\frac{a}{x}=2\Rightarrow \frac{n-3}{4}\left(\frac{a}{x}\right)=2$

$\frac{\left(4\right)}{\left(5\right)}\to \frac{n-2}{n-3}\times \frac{4}{3}=\frac{5}{3}$

$\Rightarrow 5n-15=4n-8$

$\Rightarrow n=7$

Substituting $n=7$ ins (4), $\frac{a}{x}=2\Rightarrow a=2x$

$\left(1\right)T_3{\Rightarrow }^7{C}_2{x}^{7-2}(2x{)}^2=84$

$\Rightarrow \frac{7\times 6}{2\times 1}\times x^7\times 4^2=84$

$\Rightarrow x^7=\frac{7\times 6\times 6\times 2}{7\times 6\times 2}=1$

$\Rightarrow x=1$ $a=2x=2$