If the coefficients of 2nd, 3rd and 4th terms in the expansion of ` (1+x)^(2n)` are in A.P. then

Options

A. `2n^(2) + 9n+7 = 0`
B. `2n^(2) - 9n + 7 = 0`
C. `2n^(2) - 9n - 7 = 0`
D. none of these

Answer

To solve the problem, we need to find the condition under which the coefficients of the 2nd, 3rd, and 4th terms in the expansion of \((1+x)^{2n}\) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The general term in the expansion of \((1+x)^{2n}\) is given by:
\[ T_k = \binom{2n}{k} x^k \]
Therefore, the coefficients of the 2nd, 3rd, and 4th terms are:
- 2nd term: \(T_1 = \binom{2n}{1}\)
- 3rd term: \(T_2 = \binom{2n}{2}\)
- 4th term: \(T_3 = \binom{2n}{3}\)

2. **Set Up the A.P. Condition**:
For the coefficients to be in A.P., the following condition must hold:
\[ 2 \cdot \binom{2n}{2} = \binom{2n}{1} + \binom{2n}{3} \]

3. **Substitute the Binomial Coefficients**:
We know:
- \(\binom{2n}{1} = 2n\)
- \(\binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1)\)
- \(\binom{2n}{3} = \frac{2n(2n-1)(2n-2)}{6} = \frac{2n(2n-1)(2n-2)}{6}\)

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

4. **Simplify the Equation**:
Multiply through by 6 to eliminate the fraction:
\[ 12n(2n-1) = 12n + 2n(2n-1)(2n-2) \]
Expanding both sides:
\[ 24n^2 - 12n = 12n + 2n(4n^2 - 6n + 2) \]
This simplifies to:
\[ 24n^2 - 12n = 12n + 8n^3 - 12n^2 + 4n \]

5. **Rearranging the Terms**:
Combine like terms:
\[ 8n^3 - 12n^2 + 12n - 12n = 0 \]
This simplifies to:
\[ 8n^3 - 12n^2 = 0 \]

6. **Factor the Equation**:
Factor out \(4n^2\):
\[ 4n^2(2n - 3) = 0 \]

7. **Find the Solutions**:
Setting each factor to zero gives:
\[ 4n^2 = 0 \quad \Rightarrow \quad n = 0 \]
\[ 2n - 3 = 0 \quad \Rightarrow \quad n = \frac{3}{2} \]

### Conclusion:
The values of \(n\) for which the coefficients of the 2nd, 3rd, and 4th terms in the expansion of \((1+x)^{2n}\) are in A.P. are \(n = 0\) and \(n = \frac{3}{2}\).