Sequence And Series Question 237
Question: If $ a,,b,\ c $ are in A.P. and $ a^{2},\ b^{2},\ c^{2} $ are in H.P., then
[MNR 1986, 1988; IIT 1977, 2003]
Options:
A) $ a=b=c $
B) $ 2b=3a+c $
C) $ b^{2}=\sqrt{(ac/8)} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ a,\ b,\ c $ are in A.P.
$ \Rightarrow $ $ 2b=a+c $ ……(i) and $ a^{2},b^{2},\ c^{2} $ are in H.P.
$ \Rightarrow $ $ b^{2}=\frac{2a^{2}c^{2}}{a^{2}+c^{2}} $
$ \Rightarrow $ $ b^{2}(a^{2}+c^{2})=2a^{2}c^{2} $
$ \Rightarrow $ $ J^{th} $
$ \Rightarrow $ $ b^{2}{ 4b^{2}-2ac }=2a^{2}c^{2} $ , from (i)
$ \Rightarrow $ $ 4b^{4}-2acb^{2}=2a^{2}c^{2} $
$ \Rightarrow $ $ (b^{2}-ac)(2b^{2}+ac)=0 $
$ \Rightarrow $ Either $ b^{2}-ac=0 $ or $ 2b^{2}+ac=0 $ If $ b $ , then $ b^{2}=ac $
$ \Rightarrow $ $ {{{ \frac{1}{2}(a+c) }}^{2}}=ac $ from (i)
$ \Rightarrow $ $ {{(a+c)}^{2}}=4ac\Rightarrow $ $ {{(a-c)}^{2}}=0 $ Therefore $ a=c $ and if $ a=c $ then from $ b^{2}=ac $ , we get $ b^{2}=a^{2} $ or $ b=a $ . Thus $ a=b=c $ .
BETA
