If ` ^nC_4, ^nC_5, ^nC_6` are in A.P. then the value of n is

Options

A. 7
B. 21
C. 14
D. None of these

Answer

To solve the problem where \( ^nC_4, ^nC_5, ^nC_6 \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Set Up the A.P. Condition
If \( a, b, c \) are in A.P., then the condition is:
\[ 2b = a + c \]
In our case, we can let:
- \( a = ^nC_4 \)
- \( b = ^nC_5 \)
- \( c = ^nC_6 \)

Thus, we write:
\[ 2 \cdot ^nC_5 = ^nC_4 + ^nC_6 \]

### Step 2: Write the Combinations in Terms of Factorials
Using the formula for combinations:
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]
we can express \( ^nC_4, ^nC_5, \) and \( ^nC_6 \):
\[ ^nC_4 = \frac{n!}{4!(n-4)!}, \quad ^nC_5 = \frac{n!}{5!(n-5)!}, \quad ^nC_6 = \frac{n!}{6!(n-6)!} \]

### Step 3: Substitute into the A.P. Condition
Substituting the combinations 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: Simplify the Equation
We can cancel \( n! \) from both sides (assuming \( n! \neq 0 \)):
\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

### Step 5: Rewrite Factorials
We can rewrite the factorials:
\[ 5! = 120, \quad 4! = 24, \quad 6! = 720 \]
Thus, we can express the equation as:
\[ \frac{2}{120(n-5)!} = \frac{1}{24(n-4)!} + \frac{1}{720(n-6)!} \]

### Step 6: Clear the Denominators
Multiply through by \( 720(n-5)!(n-4)!(n-6)! \) to eliminate the denominators:
\[ 2 \cdot 720 = 30(n-4)(n-5) + (n-5)(n-4) \]

### Step 7: Expand and Rearrange
Expanding the right-hand side:
\[ 1440 = 30(n^2 - 9n + 20) + (n^2 - 9n + 20) \]
Combining like terms gives:
\[ 1440 = 31(n^2 - 9n + 20) \]
This simplifies to:
\[ 31n^2 - 279n + 620 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 31, b = -279, c = 620 \):
\[ n = \frac{279 \pm \sqrt{(-279)^2 - 4 \cdot 31 \cdot 620}}{2 \cdot 31} \]

### Step 9: Calculate the Discriminant
Calculating the discriminant:
\[ (-279)^2 - 4 \cdot 31 \cdot 620 = 77841 - 77240 = 601 \]
So:
\[ n = \frac{279 \pm \sqrt{601}}{62} \]

### Step 10: Final Values
Calculating the values gives us two potential solutions for \( n \):
- \( n = 7 \)
- \( n = 14 \)

Thus, the values of \( n \) are \( 7 \) and \( 14 \).