If Cn4,Cn5  and Cn6  are in AP, the value of n  can be

∴Cn4,Cn5,Cn6  are in AP ⇒n(n−1)(n−2)(n−3)1.2.3.4,n(n−1)(n−2)(n−3)(n−4)1.2.3.4.5 ,n(n−1)(n−2)(n−3)(n−4)(n−5)1.2.3.4.5.6 Are in APDividing each by n(n−1)(n−2)(n−3)1.2.3.4 (∴n≠0,1,2,3) ∴1,(n−4)5,(n−4)(n−5)5.6 are in AP⇒2(n−4)5=1+(n−4)(n−5)5.6 ⇒12n−48=30+n2−9n+20 Or n2−21n+98=0 Or (n−14)(n−7)=0  ; n=7,14