Let X denote the number of times heads occur in n tosses of a fair coin. If $P(X=4), P(X=5)$ and $P(X=6)$ are in AP, the value of n is

7, 14

Clearly, X is a binomial variate with parameters n and $p = \frac{1}{2}$ such that

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

Now,

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

$⇒ 2P(X=5)=P(X=4)+P(X=6) 2 {^nC}_5 \left(\frac{1}{2}\right)^n= {^nC}_4 \left(\frac{1}{2}\right)^n+{^nC}_6 \left(\frac{1}{2}\right)^n$

$⇒ 2 \,\,{^nC}_5= {^nC}_4+{^nC}_6 $

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

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

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