A fair coin is tossed n times. Let x be random variable denoting the number of heads tossed. If P(x = 4), P(x = 5), P(x = 6) are in A.P. then n =

Clearly, X is a binomial variate with parameters n and p = 1/2 such that.

$P(X=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^n$

Now,

$P(X=4),P(X=5)$ and P(X = 6) are in AP.

$\begin{array}{l}\Rightarrow 2P(X=5)=P(X=4)+P(X=6)\\ 2^n{C}_5{\left(\frac{1}{2}\right)}^n{=}^n{C}_4{\left(\frac{1}{2}\right)}^n{+}^n{C}_6{\left(\frac{1}{2}\right)}^n\\ \Rightarrow 2^n{C}_5{=}^n{C}_4{+}^n{C}_6\\ \Rightarrow 2\frac{n!}{(n-5)!5!}=\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}\\ \Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}\end{array}$

$\Rightarrow n^2-21n+98=0\Rightarrow (n-7)(n-14)=0\Rightarrow n=7$ or, 14