Given that ⁿC₄, ⁿC₅ and ⁿC₆ are in A.P.

We know that for three numbers a, b, c are in A.P, the following condition holds,

⇒ 2b = a + c

So,

⇒ 2ⁿC₅ = ⁿC₄ + ⁿC₆

Adding 2ⁿC₅ on both sides we get,

⇒ 4ⁿC₅ = ⁿC₄ + ⁿC₅ + ⁿC₅ + ⁿC₆

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ 4ⁿC₅ = ^{n + 1}C₅ + ^{n + 1}C₆

⇒ 4ⁿC₅ = ^{n + 2}C₆

We know that ,

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ 24(n – 4) = n² + 2n + n + 2

⇒ 24n – 96 = n² + 3n + 2

⇒ n² – 21n + 98 = 0

⇒ n² – 14n – 7n + 98 = 0

⇒ n(n – 14) – 7(n – 14) = 0

⇒ (n – 7)(n – 14) = 0

⇒ n – 7 = 0 or n – 14 = 0

⇒ n = 7 or n = 14

∴ The values of n are 7 and 14.