If 1b−c,1c−a,1a−bbe consecutive terms of an AP, then (b−c)2,(c−a)2,(a−b)2 will be in

A
GP
B
AP
C
HP
D
none of these

Solution:

Now, we assume $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ are in AP, then we have

$\begin{array}{l} (c - a)^{2} - (b - c)^{2} = (a - b)^{2} - (c - a)^{2} \\ \Rightarrow (b - a) (2 c - a - b) = (c - b) (2 a - b - c) . . . . . . (i) \end{array}$

Also,if $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in,AP, then

$\begin{array}{l} \frac{1}{c - a} - \frac{1}{b - c} = \frac{1}{a - b} - \frac{1}{c - a} \\ \frac{b + a - 2 c}{(c - a) (b - c)} = \frac{c + b - 2 a}{(a - b) (c - a)} \\ \Rightarrow (a - b) (b + a - 2 c) = (b - c) (c + b - 2 a) \\ \Rightarrow (b - a) (2 c - a - b) = (c - b) (2 a - b - c) \end{array}$

which is equal to Eq.(i), so our hypothesis is true.

If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ be consecutive terms of an AP, then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in

Solution:

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