Question

If the 6th, 7th and 8th terms in the expansion $(x+a{)}^n$ are respectivley 112, 7 and $\frac{1}{4}$, find x,a,n.

Answer 1

It is given that,

$T_6=112,T_7=7,T_8=\frac{1}{4}$

$\therefore T_6={}^n{C}_{n-5}{x}^{n-5}\times a^5=112$

$T_7={}^n{C}_{n-6}{x}^{n-6}\times a^6=7$

and,$T_8={}^n{C}_{n-7}{x}^{n-7}\times a^7=\frac{1}{4}$

Now,

$\frac{T_7}{T_6}=\frac{{}^n{C}_{n-6}{x}^{n-6}\times a^6}{{}^n{C}_{n-5}{x}^{n-5}\times a^5}=\frac{7}{112}$

$\Rightarrow \frac{{}^n{C}_{n-6}}{{}^n{C}_{n-5}}\times \frac{a}{x}=\frac{1}{16}\Rightarrow \frac{n-6+1}{n-(n-5)+1}\times \frac{a}{x}=\frac{1}{16}$

$[\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}]\Rightarrow \frac{n-5}{6}\times \frac{a}{x}=\frac{1}{16}$

$\Rightarrow \frac{a}{x}=\frac{6}{16}\times \frac{1}{n-5}$

$\Rightarrow \frac{a}{x}=\frac{3}{8}\times \frac{1}{(n-5)}....\left(i\right)$

and,

$\frac{T_8}{T_7}=\frac{{}^n{C}_{n-7}{x}^{n-7}\times a^7}{{}^n{C}_{n-6}{x}^{n-6}\times a^6}=\frac{1}{\frac{4}{7}}$

$\Rightarrow \frac{T_8}{T_7}=\frac{{}^n{C}_{n-7}}{{}^n{C}_{n-6}}=\frac{a}{x}=\frac{1}{28}$

$\Rightarrow \frac{{}^n{C}_{n-7}}{{}^n{C}_{n-6}}\times \frac{a}{x}=\frac{1}{28}$

$\Rightarrow \frac{n-7+1}{n-(n-6)+1}\times \frac{a}{x}=\frac{1}{28}$

$\Rightarrow \frac{n-6}{7}\times \frac{a}{x}=\frac{1}{28}$

$\Rightarrow \frac{a}{x}=\frac{1}{4(n-6)}....\left(ii\right)$

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

$\frac{3}{8}\times \frac{1}{(n-5)}=\frac{1}{4(n-6)}$

$\Rightarrow \frac{3}{2}\times \frac{1}{(n-5)}=\frac{1}{(n-6)}$

$\Rightarrow 3(n-6)=2(n-5)$

$\Rightarrow 3n-18=2n-10$

$\Rightarrow 3n-2n=18-10$

$\Rightarrow n=8$

Putting in equation (i) we get

$\Rightarrow a=x$

Now, ${}^8{C}_5{x}^{8-5}{\left(\frac{x}{8}\right)}^5=112$

$=\frac{56x^8}{8^5}=112$

$\Rightarrow x^8=48$

$\Rightarrow x=4$

By putting the value of x and n in (i) we get

$a=\frac{1}{2}$

$a=\frac{1}{2}$ and x=4