If nC4,nC5 and nC6 are in A.P., then a value of n can be
6
7
8
9
As nC4,nC5 and nC6 are in A.P.
2 nC5=nC4+nC6
⇒ 2= nC4 nC5+ nC6 nC5
=n!4!(n−4)!5!(n−5)!n!+n!6!(n−6)!5!(n−5)!n!
⇒ 2=5n−4+n−56⇒n2−21n+98=0⇒ n=7,14.