A) G.P.
B) A.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
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).
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