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Show that no three consecutive binomial ...

Show that no three consecutive binomial coefficients can be in (i) G.P., (ii) H.P.

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Suppose that the r th, (r + 1) th and (r + 2) th
coefficients of ` (1 + x)^(n)` are in GP .
i.e., ` ""^(n)C_(r-1),""^(n)C_(r), ""^(n)C_(r+1) ` are in GP .
Then `(""^(n)C_(r))/(""^(n)C_(r-1))=(""^(n)C_(r+1))/(""^(n)C_(r))`
` rArr (n-r+1)/(r) = (n-r)/(r+1) [because (""^(n)C_(r))/(""^(n)C_(r-1))= (n-r+1)/(r)]`
` rArr (n-r +1)(r +1)= r (n-r)`
` rArr nr + n-r^(2) - r + r +1 = nr - r^(2)`
` rArr n+1= 0 `
` rArr n=-1 `
which is not possible , since n is a positive integer.
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