If `""^(2n)C_(1), ""^(2n)C_(2)` and `""^(2n)C_(3)` are in A.P., find n.

Answer

To solve the problem where \( \binom{2n}{1}, \binom{2n}{2}, \) and \( \binom{2n}{3} \) are in Arithmetic Progression (A.P.), we will follow these steps:

### Step 1: Write the condition for A.P.
For three numbers \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
In our case, we have:
\[ 2 \binom{2n}{2} = \binom{2n}{1} + \binom{2n}{3} \]

### Step 2: Substitute the binomial coefficients
Now we substitute the binomial coefficients:
\[ 2 \cdot \frac{(2n)!}{2!(2n-2)!} = \frac{(2n)!}{(2n-1)!} + \frac{(2n)!}{3!(2n-3)!} \]

### Step 3: Simplify the left-hand side
The left-hand side simplifies as follows:
\[ 2 \cdot \frac{(2n)!}{2!(2n-2)!} = \frac{(2n)!}{(2n-2)!} \]
This is because \( 2! = 2 \).

### Step 4: Simplify the right-hand side
Now, simplify the right-hand side:
\[ \frac{(2n)!}{(2n-1)!} + \frac{(2n)!}{3!(2n-3)!} = (2n) + \frac{(2n)!}{6(2n-3)!} \]
The second term can be simplified further:
\[ \frac{(2n)!}{6(2n-3)!} = \frac{(2n)(2n-1)(2n-2)}{6} \]

### Step 5: Equate both sides
Now we equate both sides:
\[ \frac{(2n)!}{(2n-2)!} = (2n) + \frac{(2n)(2n-1)(2n-2)}{6} \]

### Step 6: Simplify the equation
The left side can be expressed as:
\[ (2n)(2n-1) \]
Thus, we have:
\[ (2n)(2n-1) = (2n) + \frac{(2n)(2n-1)(2n-2)}{6} \]
Dividing through by \( 2n \) (assuming \( n \neq 0 \)):
\[ (2n-1) = 1 + \frac{(2n-1)(2n-2)}{6} \]

### Step 7: Clear the fraction
Multiply through by 6 to eliminate the fraction:
\[ 6(2n-1) = 6 + (2n-1)(2n-2) \]
Expanding both sides gives:
\[ 12n - 6 = 6 + (4n^2 - 6n + 2n - 2) \]
\[ 12n - 6 = 4n^2 - 4n + 4 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 4n^2 - 16n + 10 = 0 \]

### Step 9: Solving the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -16, c = 10 \):
\[ n = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 4 \cdot 10}}{2 \cdot 4} \]
\[ n = \frac{16 \pm \sqrt{256 - 160}}{8} \]
\[ n = \frac{16 \pm \sqrt{96}}{8} \]
\[ n = \frac{16 \pm 4\sqrt{6}}{8} \]
\[ n = 2 \pm \frac{\sqrt{6}}{2} \]

### Final Answer
Thus, the values of \( n \) are:
\[ n = 2 + \frac{\sqrt{6}}{2} \quad \text{or} \quad n = 2 - \frac{\sqrt{6}}{2} \]