find n if 2nc1 ,2nc2 and 2nc3 are in a.p.
Answers
Answer:
n = 7/2
Step-by-step explanation:
Given find n if 2nc1 ,2nc2 and 2nc3 are in a.p.
We know that 2nc1 ,2nc2 and 2nc3 are in a.p.
So 2 n C2 - 2 n C1 = 2 n C3 - 2 n C2
Now we know that n Cr = n! / r!(n - r)!
2 n C1 = 2 n! / 1 ! (2 n - 1)!
= 2 n x (2 n - 1)! /(2 n - 1)!
= 2 n
2 n C2 = 2 n! / 2!(2n - 2)!
= 2 n (2 n - 1)(2 n - 2)! /2 (2n - 2)!
= n(2 n - 1)
2 n C3 = 2n! /3!(2n - 3)!
= 2n(2n - 1)(2n - 2)(2n - 3)! / 3 x 2 x (2n - 3)!
= n(2n - 1)(2n - 2) / 3
Applying the concept of A.P we get
n(2n - 1) - 2n = n(2n - 1)(2n - 2) / 3 - n(2n - 1)
n(2n - 1) - 2n + n(2n - 1) = n(2n - 1)(2n - 2)/3
3(4n^2 - 4n) = 4n^3 - 6n^2 + 2n
18n^2 - 4n^3 - 14n = 0
2n(9n - 2n^2 - 7) = 0
2n(2n^2 - 7n - 2n + 7) = 0
2n (2n - 7) - 1(2n - 7) = 0
2n[ (n - 1)(2n - 7)] = 0
2n = 0, 2n - 7 = 0, n - 1 = 0
So n = 0, n = 1, n = 7/2
So n = 7/2 since it will satisfy the given equation