An unbiased coin is tossed n times. Let X denote the number of times head occurs. If `P(X=4), P(X=5) and P(X=6)` are in A.P, then the value of n can be

Options

A. 15
B. 7
C. 9
D. 14

Answer

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 (A.P.).

### Step-by-Step Solution:

1. **Understanding the Probability**:
The number of heads \( X \) in \( n \) tosses of an unbiased coin follows a binomial distribution:
\[ P(X = r) = \binom{n}{r} \left(\frac{1}{2}\right)^n \]
where \( \binom{n}{r} \) is the binomial coefficient.

2. **Setting Up the Probabilities**:
We can express the probabilities for \( X = 4, 5, 6 \):
\[ P(X = 4) = \binom{n}{4} \left(\frac{1}{2}\right)^n \]
\[ P(X = 5) = \binom{n}{5} \left(\frac{1}{2}\right)^n \]
\[ P(X = 6) = \binom{n}{6} \left(\frac{1}{2}\right)^n \]

3. **Condition for A.P.**:
The condition for three numbers \( a, b, c \) to be in A.P. is:
\[ 2b = a + c \]
Applying this to our probabilities:
\[ 2P(X = 5) = P(X = 4) + P(X = 6) \]
Substituting the probabilities:
\[ 2 \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{4} \left(\frac{1}{2}\right)^n + \binom{n}{6} \left(\frac{1}{2}\right)^n \]
Dividing through by \( \left(\frac{1}{2}\right)^n \) (which is non-zero):
\[ 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

4. **Using the Binomial Coefficient Identity**:
We can use the identity:
\[ \binom{n}{r} = \frac{n-r+1}{r} \binom{n}{r-1} \]
This gives us:
\[ \binom{n}{4} = \frac{n-4}{4} \binom{n}{3}, \quad \binom{n}{5} = \frac{n-5}{5} \binom{n}{4}, \quad \binom{n}{6} = \frac{n-6}{6} \binom{n}{5} \]

5. **Substituting the Values**:
Substituting back into our equation:
\[ 2 \cdot \frac{n-5}{5} \binom{n}{4} = \binom{n}{4} + \frac{n-6}{6} \cdot \frac{n-5}{5} \binom{n}{4} \]
Factoring out \( \binom{n}{4} \):
\[ 2 \cdot \frac{n-5}{5} = 1 + \frac{(n-6)(n-5)}{30} \]

6. **Clearing the Denominators**:
Multiply through by \( 30 \):
\[ 12(n-5) = 30 + (n-6)(n-5) \]
Expanding:
\[ 12n - 60 = 30 + n^2 - 11n + 30 \]
Rearranging:
\[ n^2 - 23n + 120 = 0 \]

7. **Solving the Quadratic Equation**:
We can factor or use the quadratic formula:
\[ n = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 1 \cdot 120}}{2 \cdot 1} \]
\[ n = \frac{23 \pm \sqrt{529 - 480}}{2} \]
\[ n = \frac{23 \pm \sqrt{49}}{2} \]
\[ n = \frac{23 \pm 7}{2} \]
This gives:
\[ n = \frac{30}{2} = 15 \quad \text{or} \quad n = \frac{16}{2} = 8 \]

8. **Final Values**:
Therefore, the possible values of \( n \) are \( 15 \) and \( 8 \).

### Conclusion:
The values of \( n \) can be \( 8 \) and \( 15 \).