If the coefficients of $5\text{th}$ , $\text{6th}$  and $\text{7th}$  terms in the expansion of ${(1+x)}^n,n\in N$ , are in A.P., then $n$  is equal to

The coefficients of $x^{r-1},x^r,x^{r+1}$  in the expansion of ${(1+x)}^n$  are

${ }^n{C}_{r-1},{}^n{C}_r,{}^n{C}_{r+1}$  Respectively. Since these are in A.P. we have (suppose a,b,c are in AP

 then $2b=a+c$ ) here $a={}^n{C}_{r-1},b={}^n{C}_r,c={}^n{C}_{r+1}$  then

  $2^n{C}_r={}^2{C}_{r-1}+{}^n{C}_{r+1}$  wkt ${ }^n{C}_r=\frac{\overline{)n}}{\overline{)r}\overline{)n-r}}$

$\Rightarrow \frac{2\overline{)n}}{\overline{)r}\overline{)n-r}}=\frac{\overline{)n}}{\overline{)r-1}\overline{)n-r+1}}+\frac{\overline{)n}}{\overline{)r+1}\overline{)n-r-1}}$

$\Rightarrow \frac{2\overline{)n}}{r\overline{)r-1}\left(n-r\right)\overline{)n-r-1}}=\frac{\overline{)n}}{\overline{)r-1}(n-r+1)\left(n-r\right)\overline{)n-r-1}}+\frac{\overline{)n}}{r(r+1)\overline{)r-1}\overline{)n-r-1}}$

(Both sides cancellation of terms $\overline{)n}$ ,$\overline{)r-1}$ ,$\overline{)n-r-1}$  then we get)

$\Rightarrow \frac{2}{r(n-r)}=\frac{1}{(n-r+1)(n-r)}+\frac{1}{(r+1)r}$

$\begin{array}{c}\Rightarrow \frac{2}{r(n-r)}=\frac{1}{(n-r+1)(n-r)}+\frac{1}{(r+1)r}\\ \Rightarrow \frac{2}{r(n-r)}=\frac{(r+1)r+(n-r+1)(n-r)}{(n-r+1)(n-r)(r+1)r}\end{array}$

(Both sides cancellation of terms,$r(n-r)$  then we get)

$\begin{array}{c} \\ \Rightarrow \frac{2}{1}=\frac{(r+1)r+(n-r+1)(n-r)}{(n-r+1)(r+1)}\end{array}$

Now cross multiplication we get

  $2(n-r+1)(r+1)=r(r+1)+(n-r+1)(n-r)$

$\begin{array}{c}\Rightarrow 2(n-r+1)(r+1)=r(r+1)+(n-r+1)(n-r)\\ \Rightarrow 2(nr+n-r^2-r+r+1)=r^2+r+n^2-nr-rn+r^2+n-r\end{array}$

$\Rightarrow n^2-n(4r+1)+4r^2-2=0$

Deduction. If the coefficients of $5\text{th}$ ,$\text{6th}$  and $\text{7th}$  terms in the expansion of

${(1+x)}^n$  are in A.P. then ${ }^n{C}_4{,}^n{C}_5{,}^n{C}_6$

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

$\begin{array}{c}\Rightarrow n^2-(4.5+1)n+4\left(25\right)-2=0\Rightarrow n^2-21n+98=0\\ \Rightarrow n=7,14\end{array}$  

Solving the above Eqn  we get  7,14