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

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 2nC2nC2 2nC3 are in A.P.. 

Therefore, (2nC2) / (2nC1) = (2nC3) / (2nC2)

ie. 2nC2 × 2nC2 = 2nC3 × 2nC1  

ie.  2n! ÷ 2!(2n-2)!  × 2n! ÷ 2!(2n-2)! = 2n! ÷ 3!(2n-3)! × 2n! ÷ 1!(2n-1)!

ie.2n(2n-1)(2n-2)! ÷ 2!(2n-2)! × 2n(2n-1)(2n-2)! ÷ 2!(2n-2)! = 2n(2n-1)(2n-2)(2n-3)! ÷ 3!(2n-3)! × 2n(2n-1)÷ 1!(2n-1)!

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