Question

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

Answer 1

It is given that

$T_2=240$

$T_3=720$

$T_4=1080$

$\therefore T_2={}^n{C}_1\times x^{n-1}\times a=240$

$T_3={}^n{C}_2\times x^{n-2}\times a^2=720$

$T_4={}^n{C}_3\times x^{n-3}\times a^3=1080$

Now,

$\frac{T_4}{T_3}=\frac{{}^n{C}_3\times x^{n-3}\times a^3}{{}^n{C}_2\times n^{n-2}\times a^2}=\frac{1080}{720}$

$\Rightarrow \frac{{}^n{C}_{3}a}{{}^n{C}_{2}x}=\frac{3}{2}$

$\Rightarrow \frac{n-3+1}{2+1}\times \frac{a}{x}=\frac{3}{2}$

$\Rightarrow \frac{n-2}{3}\times \frac{a}{x}=\frac{3}{2}$

$\Rightarrow \frac{a}{x}=\frac{9}{2(n-2)}.......\left(i\right)$

and,

$\frac{T_3}{T_2}=\frac{{}^n{C}_2\times x^{n-2}\times a^2}{{}^n{C}_1\times x^{n-1}\times a}=\frac{720}{240}$

$\Rightarrow \frac{{}^n{C}_2}{{}^n{C}_1}\times \frac{a}{x}=3$

$\Rightarrow \frac{n-2+1}{2}\times \frac{a}{x}=3$

$\Rightarrow \frac{n-1}{2}\times \frac{a}{x}=3$

$\Rightarrow \frac{a}{x}=\frac{6}{n-1}.....\left(ii\right)$

Comparing equation (i) and (ii), we get

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

$\Rightarrow 12(n-2)=9(n-1)$

$\Rightarrow 12n-24=9n-9$

$\Rightarrow 3n=24-9$

$\Rightarrow 3n=15$

$\Rightarrow n=5$

Putting n =5 in equation (ii), we get

$\frac{a}{x}=\frac{6}{5-1}$

$\Rightarrow \frac{a}{x}=\frac{6}{4}$

$\Rightarrow \frac{a}{x}=\frac{3}{2}$

$\Rightarrow a=\frac{3}{2}x$

Now,

\)T_2 ={^nC_1} \times x^{n-1}\times a=240\)

${\Rightarrow }^5{C}_1\times x^4\times \left(\frac{3}{2}x\right)=240$

$[\because n=5anda=\frac{3}{2}x]$

$\Rightarrow x^5=\frac{240\times 2}{5\times 3}$

$\Rightarrow x^5=32$

$\Rightarrow x^5=2^5$

$\Rightarrow x^5=2$

Putting x =2 in $a=\frac{3}{2}x$, we get

$a=\frac{3}{2}\times 2=3$

Hence, x =2, a =3 and n =5.