Find n if ²ⁿC₁ ,2n c  ₂  and ²ⁿC₃ are in A.P?

 ²ⁿC₁ = 2n                         ²ⁿC₂ = (2n)!/2!(2n-2)! = n(2n-1)           ²ⁿC₃ = (2n)!/3!(2n-3)! = 2n(2n-1)(n-1)/3

If they are in A.P ;   ²ⁿC₂ - ²ⁿC₁  =  ²ⁿC₃ - ²ⁿC₂

 n(2n-1) - 2n  = 2n(2n-1)(n-1)/3 - n(2n-1)

n(2n-3) =  [2n(2n-1)(n-1) - 3n(2n-1)] /3

n(2n-3) = n(2n-1)(2n-5) / 3

(2n-3) = (2n-1)(2n-5) / 3

6n-9 = 4n²-10n-2n+5

4n²-18n+14 = 0

2n²-9n+7 = 0

n = 9 +/- (81-59)^1/2 / 4 = 9 +/- 5 /4 = 7/2 or 1

n can't be 1 bcoz when n=1 ²ⁿC₃ will be ²C₃ which is not possible. so n = 7/2