Step 1: Set up
Let the common ratio idea be avoided. Write the three coefficients with factorials relative to ${}^nC_4$.
Step 2: Express
${}^nC_5 = {}^nC_4 \cdot \frac{n-4}{5}$ and ${}^nC_6 = {}^nC_5 \cdot \frac{n-5}{6}$.
Step 3: Use the AP
$2\,{}^nC_5 - {}^nC_4 - {}^nC_6 = 0$ leads to $2 = \frac{5}{n-4} + \frac{n-5}{6}$ after dividing by ${}^nC_5$. This simplifies to $n^2 - 21n + 98 = 0$.
Step 4: Test the options
The roots are 7 and 14. Check $n = 14$: ${}^{14}C_4 = 1001$, ${}^{14}C_5 = 2002$, ${}^{14}C_6 = 3003$, which are in A.P. with difference 1001. The other pairs (5 and 11, 8 and 15, 6 and 13) do not satisfy the quadratic.
Final Answer:
n equals 7 or 14. This is option (B).
\[ \boxed{\text{(B) }7 \text{ or } 14} \]