Show that no three consecutive binomial coefficients can be in G.P.

Explanation

Let `.^(n)C_(r-1), .^(n)C_(r),.^(n)C_(r+1)` be in G.P.
`:. (.^(n)C_(r))/(.^(n)C_(r-1)) = (.^(n)C_(r+1))/(.^(n)C_(r))`
or `(n-r+1)/(r) = (n-(r+1)+1)/(r+1)`
or `nr-r^(2)+r+n-r+1=nr-r^(2)`
or `n + 1 =0`
or `n = - 1`
This is not possible .
Hence not in G.P.
(b) Let `.^(n)C_(r-1),.^(n)C_(r),.^(n)C_(r+1)` be in H.P.
`:. (1)/(.^(n)C_(r-1)),(1)/(.^(n)C_(r)),.(1)/(.^(n)C_(r+1))` are in A.P.
`rArr (2)/(.^(n)C_(r))=(1)/(.^(n)C_(r-1))+(1)/(.^(n)C_(r+1))`
or `2 = (.^(n)C_(r))/(.^(n)C_(r-1))+(.^(n)C_(r))/(.^(n)C_(r+1))`
or `2 = (n-r+1)/(r)+(r+1)/(n-r)`
or `2r(n-r)=(n-r)^(2)+(n-r)+r^(2)+r`
or `2rn-2r^(2)=n^(2)-2nr+r^(2)+n-r+r^(2)+r`
or `n^(2)+4r^(2)-4nr+n=0`
or `(n-2r)^(2) +n = 0`
This is not possible as both `(n-2r)^(n)` and n are positive.
Hence not in H.P.