Mathematics - Binomial Theorem Question with Solution | TestHub

$a_{n-3}=a_3,a_{n-2}=a_2,a_{n-1}=a_1\left(\because {}^n{C}_r={}^n{C}_{n-r}\right)$ $\begin{array}{rlrlrlr}& \therefore a_1,a_2,a_3\text{ are in AP }& \Rightarrow n,\frac{n(n-1)}{2},\frac{n(n-1)(n-2)}{6}\text{ are in AP. }& \frac{n+\frac{n(n-1)(n-2)}{6}}{2}=\frac{n(n-1)}{2}& 6n+n^3-3n^2+2n=6n^2-6n& n^3-9n^2+14n=0,(n-7)(n-2)=0& \therefore \mathrm{n}=7.\end{array}$