If the coefficients `""^nC_4, ""^nC_5, ""^nC_6` of `(1 +x)^n` are in A.P. then n is equal to

Options

A. 12
B. 11
C. 7
D. 8

Answer

To solve the problem, we need to determine the value of \( n \) such that the coefficients \( \binom{n}{4} \), \( \binom{n}{5} \), and \( \binom{n}{6} \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Condition for A.P.**:
The coefficients \( \binom{n}{4} \), \( \binom{n}{5} \), and \( \binom{n}{6} \) are in A.P. if:
\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

2. **Using the Binomial Coefficient Formula**:
The binomial coefficients can be expressed as:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Therefore, we can write:
\[ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]

3. **Substituting the Coefficients**:
Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]

4. **Simplifying the Equation**:
Dividing through by \( n! \) (assuming \( n \geq 6 \)):
\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

5. **Finding a Common Denominator**:
The common denominator for the right side is \( 4! \cdot 6! \cdot (n-4)! \cdot (n-6)! \). Rewriting gives:
\[ 2 \cdot (n-4)! = \frac{(n-4)! \cdot (n-6)!}{5!} + \frac{(n-4)! \cdot (n-5)!}{6!} \]

6. **Cross Multiplying**:
This leads to:
\[ 2 \cdot 6! \cdot (n-4)! = (n-5)! + (n-6)! \]

7. **Substituting Values**:
Let \( r = 5 \) (as \( r \) is the middle index):
\[ n - 2r = n - 10 \]
Therefore, we set up the equation:
\[ (n - 10)^2 = n + 2 \]

8. **Expanding and Rearranging**:
Expanding gives:
\[ n^2 - 20n + 100 = n + 2 \]
Rearranging leads to:
\[ n^2 - 21n + 98 = 0 \]

9. **Factoring the Quadratic**:
Factoring gives:
\[ (n - 14)(n - 7) = 0 \]
Hence, \( n = 14 \) or \( n = 7 \).

10. **Conclusion**:
Since the problem states that the coefficients are in A.P., we check for valid values. The only valid solution is:
\[ n = 7 \]

### Final Answer:
Thus, the value of \( n \) is \( 7 \).