Find 'n', if `""^(2n)C_(1), ""^(2n)C_(2)` and `""^(2n)C_(3)` are in A.P.

Answer

To solve the problem of finding \( n \) such that \( \binom{2n}{1}, \binom{2n}{2}, \binom{2n}{3} \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Set up 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 can set:
- \( a = \binom{2n}{1} \)
- \( b = \binom{2n}{2} \)
- \( c = \binom{2n}{3} \)

Thus, we have:
\[ 2 \binom{2n}{2} = \binom{2n}{1} + \binom{2n}{3} \]

### Step 2: Substitute the binomial coefficients
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
we can write:
- \( \binom{2n}{1} = \frac{2n!}{1!(2n-1)!} = 2n \)
- \( \binom{2n}{2} = \frac{2n!}{2!(2n-2)!} = \frac{2n(2n-1)}{2} = n(2n-1) \)
- \( \binom{2n}{3} = \frac{2n!}{3!(2n-3)!} = \frac{2n(2n-1)(2n-2)}{3!} = \frac{2n(2n-1)(2n-2)}{6} \)

### Step 3: Substitute into the A.P. condition
Substituting these into the A.P. condition gives:
\[ 2(n(2n-1)) = 2n + \frac{2n(2n-1)(2n-2)}{6} \]

### Step 4: Simplify the equation
Multiply through by 6 to eliminate the fraction:
\[ 12n(2n-1) = 12n + 2n(2n-1)(2n-2) \]
Expanding the right-hand side:
\[ 12n(2n-1) = 12n + 2n(2n-1)(2n-2) \]
This simplifies to:
\[ 12n(2n-1) = 12n + 2n(4n^2 - 6n + 2) \]

### Step 5: Rearranging and simplifying
Rearranging gives:
\[ 12n(2n-1) - 12n = 2n(4n^2 - 6n + 2) \]
Factoring out \( 2n \):
\[ 2n(6(2n-1) - 6) = 2n(4n^2 - 6n + 2) \]
Now, we can cancel \( 2n \) (assuming \( n \neq 0 \)):
\[ 6(2n-1) - 6 = 4n^2 - 6n + 2 \]

### Step 6: Solve the resulting equation
This simplifies to:
\[ 12n - 6 - 6 = 4n^2 - 6n + 2 \]
\[ 12n - 12 = 4n^2 - 6n + 2 \]
Rearranging gives:
\[ 4n^2 - 18n + 14 = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 4, b = -18, c = 14 \):
\[ n = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 4 \cdot 14}}{2 \cdot 4} \]
Calculating the discriminant:
\[ n = \frac{18 \pm \sqrt{324 - 224}}{8} \]
\[ n = \frac{18 \pm \sqrt{100}}{8} \]
\[ n = \frac{18 \pm 10}{8} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{28}{8} = 3.5 \)
2. \( n = \frac{8}{8} = 1 \)

### Final Answer
Thus, the possible values of \( n \) are \( n = 1 \) or \( n = 3.5 \).