Question

The coefficients of 5th, 6th and 7th terms in the expansion of $(1+x{)}^n$ are in A.P., find n.

Answer 1

We have,

$(1+x{)}^n$

Now,

Coefficient of 5th term $={}^n{C}_{5-1}={}^n{C}_4$

Coefficient of 5th term $={}^n{C}_{6-1}={}^n{C}_5$ and, Coefficient of 5th term $={}^n{C}_{7-1}={}^n{C}_6$

It is given that these coefficients are in A.P.

$\therefore {}^{2n}{C}_5={}^n{C}_4+{}^n{C}_6$

$\Rightarrow 2\left[\frac{n!}{(n-5)!5!}\right]=\frac{n!}{(n-r)!4!}+\frac{n!}{(n-6)!6!}$

$\Rightarrow \frac{2}{(n-5)!5!}=\frac{1}{(n-4)!4!}+\frac{1}{(n-6)!6!}$

$\Rightarrow \frac{2}{(n-5)(n-6)!5\times 4!}$

$=\frac{1}{(n-4)(n-5)(n-6)!4}+\frac{1}{(n-6)!6\times 5\times 4!}$

$\Rightarrow \frac{2}{(n-5)\times 5}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}$

$\Rightarrow \frac{2}{5(x-5)}-\frac{1}{30}=\frac{1}{(n-4)(n-5)}$

$\Rightarrow \frac{12-(n-5)}{30(x-5)}=\frac{1}{(n-4)(n-5)}$

$\Rightarrow \frac{12-n+5}{30}=\frac{1}{(n-4)(n-5)}$

$\Rightarrow \frac{17-n}{30}=\frac{1}{n-4}$

$\Rightarrow 17n-68-n^2+4n=30$

$\Rightarrow 21n-68-m^2-30=0$

$\Rightarrow 21n-n^2-98=0$

$\Rightarrow n^2-21n+98=0$

$\Rightarrow n^{2}7n-14n+98=0$

$\Rightarrow n(n-7)-17(n-7)=0$

$\Rightarrow (n-7)(n-14)=0$

$\Rightarrow n=7$ or n =14