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 a. `7, 14` b. `10 , 14` c. `12 ,7` d. `14 , 12`

Explanation

To solve the problem, we need to find the value of \( n \) such that the probabilities \( P(X=4) \), \( P(X=5) \), and \( P(X=6) \) are in arithmetic progression (AP).

### Step-by-step Solution:

1. **Understanding the Binomial Distribution**:
The number of heads \( X \) in \( n \) tosses of a fair coin follows a binomial distribution with parameters \( n \) and \( p = \frac{1}{2} \). The probability mass function is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} = \binom{n}{k} \left(\frac{1}{2}\right)^n
...