If frac{1}{{b - c}},;frac{1}{{c - a}},;frac{1 | vedclass

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Question

(b) If ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ are in $A.P.$

Then we have ${(c - a)^2} - {(b - c)^2} = {(a - b)^2} - {(c - a)^2}$

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$A.P.$

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(b) If ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ are in $A.P.$

Then we have ${(c - a)^2} - {(b - c)^2} = {(a - b)^2} - {(c - a)^2}$

$ \Rightarrow $ $(b - a)(2c - a - b) = (c - b)(2a - b - c)$&hellip;..(i)

Also if $\frac{1}{{b - c}},\;\frac{1}{{c - a}},\frac{1}{{a - b}}$are in $A.P.$

Then $\frac{1}{{c - a}} - \frac{1}{{b - c}} = \frac{1}{{a - b}} - \frac{1}{{c - a}}$

$ \Rightarrow $ $\frac{{b + a - 2c}}{{(c - a)(b - c)}} = \frac{{c + b - 2a}}{{(a - b)(c - a)}}$

$ \Rightarrow $ $(a - b)(b + a - 2c) = (b - c)(c + b - 2a)$

$ \Rightarrow $ $ (b - a)(2c - a - b) = (c - b)(2a - b - c)$,

which is true by virtue of $(i).$

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

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