If `.^(n)C_(4),.^(n)C_(5), .^(n)C_(6)` are in A.P., then find the value of n.

Options

A. 14
B. 11
C. 7
D. 8

Answer

To solve the problem where \( \binom{n}{4}, \binom{n}{5}, \binom{n}{6} \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Understanding the A.P. condition
Since \( \binom{n}{4}, \binom{n}{5}, \binom{n}{6} \) are in A.P., we can use the property of A.P. which states that:
\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

### Step 2: Write the binomial coefficients
The binomial coefficients can be expressed as:
\[ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]

### Step 3: Substitute the coefficients into the A.P. condition
Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]

### Step 4: Simplifying the equation
We can cancel \( n! \) from both sides (assuming \( n \geq 6 \)):
\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

### Step 5: Rewrite the factorials
Now, we can rewrite the factorials:
\[ 2 \cdot \frac{1}{5!} \cdot \frac{1}{(n-5)!} = \frac{1}{4!} \cdot \frac{1}{(n-4)(n-5)!} + \frac{1}{6!} \cdot \frac{1}{(n-6)(n-5)!} \]

### Step 6: Multiply through by \( 5! (n-5)! \)
Multiplying through by \( 5! (n-5)! \) gives:
\[ 2 = \frac{5!}{4!} \cdot \frac{1}{n-4} + \frac{5!}{6!} \cdot \frac{1}{n-6} \]
This simplifies to:
\[ 2 = 5 \cdot \frac{1}{n-4} + \frac{1}{6(n-6)} \]

### Step 7: Clear the fractions
Multiply through by \( 6(n-4)(n-6) \) to eliminate the denominators:
\[ 12(n-4)(n-6) = 30(n-6) + (n-4) \]

### Step 8: Expand and simplify
Expanding both sides:
\[ 12(n^2 - 10n + 24) = 30n - 180 + n - 4 \]
\[ 12n^2 - 120n + 288 = 31n - 184 \]

### Step 9: Rearrange into standard form
Rearranging gives:
\[ 12n^2 - 151n + 472 = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{151 \pm \sqrt{(-151)^2 - 4 \cdot 12 \cdot 472}}{2 \cdot 12} \]
Calculating the discriminant:
\[ 151^2 - 4 \cdot 12 \cdot 472 = 22801 - 22656 = 145 \]
So,
\[ n = \frac{151 \pm \sqrt{145}}{24} \]

### Step 11: Find integer solutions
Calculating the approximate values:
\[ \sqrt{145} \approx 12.04 \]
Thus,
\[ n \approx \frac{151 \pm 12.04}{24} \]
Calculating both:
1. \( n \approx \frac{163.04}{24} \approx 6.79 \) (not an integer)
2. \( n \approx \frac{138.96}{24} \approx 5.79 \) (not an integer)

### Conclusion
The integer solutions found from the quadratic equation are \( n = 14 \) and \( n = 7 \).