The problem involves determining the number of times heads appear when a fair coin is tossed \( n \) times, where the probabilities \( P(X=4) \), \( P(X=5) \), and \( P(X=6) \) are in an arithmetic progression (AP). We need to find the value of \( n \).
Given,
Let's recall the formula for binomial probability:
\(P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n\)
So, we employ it here:
\[ 2 \times \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{4} \left(\frac{1}{2}\right)^n + \binom{n}{6} \left(\frac{1}{2}\right)^n \]
Simplify this equation by multiplying all terms by \( (2^n) \):
\[ 2 \times \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]
Use the identities for combinations:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Plugging in the values:
\[ 2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]
Cancel \( n! \) part and simplify:
\[ 2 \times \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]
Rearranging and simplifying:
\[ 2 \times \frac{(n-4)!}{5!} = \frac{(n-5)!}{4!} + \frac{(n-6)!}{6!} \]
After substitution and other arithmetic simplifications, we find:
The equation resolves to satisfy when \( n = 7 \) and \( n = 14 \).
Thus, the correct option is 7, 14.