A coin tossed $n$ times. If the probability that $4, 5, 6$ heads occur are in $A.P.$, then $n =$

Probabilities of k heads: P(k) = C(n, k) * (1/2)^n. If P(4), P(5), P(6) are in AP, then 2 * P(5) = P(4) + P(6). This simplifies to 2 * C(n, 5) = C(n, 4) + C(n, 6). Using the formula C(n, k) = n! / (k!(n-k)!), we solve for n: 2 * (n! / (5!(n-5)!)) = n! / (4!(n-4)!) + n! / (6!(n-6)!). Dividing by n! and simplifying leads to n = 14.