If the coefficients of 2nd , 3rd and 4th terms of the expansion of `(1+x)^(2n)` are in A.P. then the value of `2n^2 -9n + 7` is

Options

A. 0
B. 5
C. 2
D. 6

Answer

To solve the problem, we need to find the value of \(2n^2 - 9n + 7\) given that the coefficients of the 2nd, 3rd, and 4th terms of the expansion of \((1+x)^{2n}\) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficients of the \(r\)-th term in the expansion of \((1+x)^{2n}\) is given by \(\binom{2n}{r-1}\). Therefore, the coefficients of the 2nd, 3rd, and 4th terms are:
- 2nd term: \(\binom{2n}{1}\)
- 3rd term: \(\binom{2n}{2}\)
- 4th term: \(\binom{2n}{3}\)

2. **Set Up the A.P. Condition**:
Since these coefficients are in A.P., we can write the condition:
\[ 2 \cdot \binom{2n}{2} = \binom{2n}{1} + \binom{2n}{3} \]

3. **Substitute the Binomial Coefficients**:
We know:
\[ \binom{2n}{1} = 2n, \quad \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1), \quad \binom{2n}{3} = \frac{2n(2n-1)(2n-2)}{6} \]
Substituting these into the A.P. condition gives:
\[ 2 \cdot n(2n-1) = 2n + \frac{2n(2n-1)(2n-2)}{6} \]

4. **Simplify the Equation**:
Multiply through by 6 to eliminate the fraction:
\[ 12n(2n-1) = 12n + 2n(2n-1)(2n-2) \]
Expanding the right side:
\[ 12n(2n-1) = 12n + 2n(2n-1)(2n-2) \]
This simplifies to:
\[ 12n(2n-1) - 12n = 2n(2n-1)(2n-2) \]
\[ 12n(2n-1 - 1) = 2n(2n-1)(2n-2) \]
\[ 12n(2n-2) = 2n(2n-1)(2n-2) \]

5. **Cancel Common Terms**:
Assuming \(n \neq 0\) (since \(n=0\) leads to trivial solutions), we can divide both sides by \(2n(2n-2)\):
\[ 6 = (2n-1) \]
Thus:
\[ 2n - 1 = 6 \implies 2n = 7 \implies n = \frac{7}{2} \]

6. **Substitute \(n\) into the Expression**:
Now, substitute \(n = \frac{7}{2}\) into \(2n^2 - 9n + 7\):
\[ 2\left(\frac{7}{2}\right)^2 - 9\left(\frac{7}{2}\right) + 7 \]
\[ = 2 \cdot \frac{49}{4} - \frac{63}{2} + 7 \]
\[ = \frac{98}{4} - \frac{126}{4} + \frac{28}{4} \]
\[ = \frac{98 - 126 + 28}{4} = \frac{0}{4} = 0 \]

### Final Answer:
Thus, the value of \(2n^2 - 9n + 7\) is \(0\).