Three consecutive binomial coefficients can never be in

Let Tr,Tr+1,Tr+2  be in G.P⇒TrTr+1,1,Tr+2Tr+1 are in G.P.  ⇒rn−r+1,1,n−rr+1are in G.P. ⇒12=r(n−r)(n−r+1)(r+1) ⇒n(r+1)−(r2−1)=nr−r2 ⇒n+1=0 ⇒n=−1,which is not possible.Again if,Tr,Tr+1,Tr+2 are in H.P ⇒1Tr,1Tr+1,1Tr+2are in A.P. ⇒ Tr+1Tr,1,Tr+1Tr+2are in A.P. ⇒ 2=n−r+1r+r+1n−r ⇒2r(n−r)=(n−r)2+(n−r)+r2+r ⇒2rn−2r2=n2−2nr+r2+n−r+r2+r ⇒n2+4r2−4nr+n=0⇒(n−2r)2+n=0This is not possible as both (n – 2r)2 and n are positive.