Sequence And Series Question 195
Question: 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
Options:
A) G.P.
B) A.P.
C) H.P.
D) None of these
Show Answer
Answer:
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).
BETA
