If n C _ 4 , ^ n C _ 5 and n C _ 6 are in A.P., then n can be

Since, $n{C}_{4}, {}^{n}{C}_{5}, {}^{n}{C}_{6}$ are in A.P.

$⇒ 2{\times}^{n}{C}_{5}{=}^{n}{C}_{4}{+}^{n}{C}_{6}$

$⇒ 2 \times \dfrac{n !}{5 ! \left(n - 5\right) !} = \dfrac{n !}{4 ! \left(n - 4\right) !} + \dfrac{n !}{6 ! \left(n - 6\right) !}$

$⇒ \dfrac{2}{5 ! \left(n - 5\right) !} = \dfrac{30 + \left(n - 5\right)\left(n - 4\right)}{6 ! \left(n - 4\right) !}$

$⇒ 6 \times 2\left(n - 4\right) = 30 + {n}^{2} - 9n + 20$

$⇒ {n}^{2} - 21n + 98 = 0$

$⇒ \left(n - 7\right)\left(n - 14\right) = 0$

$⇒ n = 7$ or $14$