If 2nC4,2nC5 and 2nC6 are in A.P, then n is equal to
14
12
7
6
2 2nC5=2nC4+2nC6⇒2= 2nC4 2nC5+ 2nC62n=52n−4+2n−56⇒12(2n−4)=30+(2n−4)(2n−5)⇒2n2−21n+49=0⇒n=7