First slide

Binomial theorem for positive integral Index

Question

Three consecutive binomial coefficients can never be in

Moderate

A

G.P
B

H.P
C

A.P
D

A.G.P

Solution

Let $Tr, {Tr}_{+ 1}, {Tr}_{+ 2}$ be in G.P
$\Rightarrow \frac{T_{r}}{T_{r + 1}}, 1, \frac{T_{r + 2}}{T_{r + 1}} are in G.P.$
$\begin{aligned} \Rightarrow \frac{r}{n - r + 1}, 1, \frac{n - r}{r + 1} are in G.P. \\ \Rightarrow 1^{2} = \frac{r (n - r)}{(n - r + 1) (r + 1)} \\ \Rightarrow n (r + 1) - (r^{2} - 1) = nr - r^{2} \\ \Rightarrow n + 1 = 0 \\ \Rightarrow n = - 1, which is not possible. \end{aligned}$
Again if, $Tr, {Tr}_{+ 1}, {Tr}_{+ 2}$ are in H.P
$\begin{aligned} \Rightarrow \frac{1}{T_{r}}, \frac{1}{T_{r + 1}}, \frac{1}{T_{r + 2}} are in A.P. \\ \Rightarrow \frac{T_{r + 1}}{T_{r}}, 1, \frac{T_{r + 1}}{T_{r + 2}} are in A.P. \\ \Rightarrow 2 = \frac{n - r + 1}{r} + \frac{r + 1}{n - r} \end{aligned}$
$\begin{aligned} \Rightarrow 2 r (n - r) = (n - r)^{2} + (n - r) + r^{2} + r \\ \Rightarrow 2 rn - 2 r^{2} = n^{2} - 2 nr + r^{2} + n - r + r^{2} + r \\ \Rightarrow n^{2} + 4 r^{2} - 4 nr + n = 0 \Rightarrow (n - 2 r)^{2} + n = 0 \end{aligned}$
This is not possible as both (n – 2r)² and n are positive.

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