If the coefficients of second, third and fourth 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 second, third, and fourth terms in the expansion of \((1 + x)^{2n}\) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficients of the second, third, and fourth terms in the expansion of \((1 + x)^{2n}\) can be expressed using the binomial coefficient:
- Coefficient of the 2nd term: \( T_2 = \binom{2n}{1} = 2n \)
- Coefficient of the 3rd term: \( T_3 = \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1) \)
- Coefficient of the 4th term: \( T_4 = \binom{2n}{3} = \frac{2n(2n-1)(2n-2)}{6} = \frac{n(2n-1)(2n-2)}{3} \)

2. **Set Up the A.P. Condition**:
The condition for three numbers \(a\), \(b\), and \(c\) to be in A.P. is given by:
\[ 2b = a + c \]
Applying this to our coefficients:
\[ 2 \cdot n(2n-1) = 2n + \frac{n(2n-1)(2n-2)}{3} \]

3. **Simplify the Equation**:
Multiply through by 3 to eliminate the fraction:
\[ 6n(2n-1) = 6n + n(2n-1)(2n-2) \]
Expanding the right-hand side:
\[ 6n(2n-1) = 6n + n(2n-1)(2n-2) \]
\[ 6n(2n-1) = 6n + n(2n^2 - 6n + 4) \]

4. **Rearranging the Equation**:
Move all terms to one side:
\[ 6n(2n-1) - 6n - n(2n^2 - 6n + 4) = 0 \]
Simplifying this gives:
\[ 12n^2 - 6n - 6n - 2n^3 + 6n^2 - 4n = 0 \]
Combine like terms:
\[ -2n^3 + 18n^2 - 16n = 0 \]

5. **Factor the Equation**:
Factor out \(n\):
\[ n(-2n^2 + 18n - 16) = 0 \]
This gives us \(n = 0\) or solving the quadratic:
\[ -2n^2 + 18n - 16 = 0 \]
Dividing through by -2:
\[ n^2 - 9n + 8 = 0 \]

6. **Solve the Quadratic**:
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2} \]
This gives us:
\[ n = 8 \quad \text{or} \quad n = 1 \]

### Conclusion:
The values of \(n\) for which the coefficients of the second, third, and fourth terms in the expansion of \((1 + x)^{2n}\) are in A.P. are \(n = 1\) and \(n = 8\).