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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
$G.P.$
$A.P.$
$H.P.$
None of these
Solution
(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 $ $(b – a)(2c – a – b) = (c – b)(2a – b – c)$…..(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 $(i).$
