Sequence And Series Question 219
Question: If $ a,\ b,\ c $ are in G.P., $ a-b,\ c-a,\ b-c $ are in H.P., then $ a+4b+c $ is equal to
Options:
A) 0
B) $ 1 $
C) $ -1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
As given $ b^{2}=ac $ and $ \frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}=\frac{a-c}{(a-b)(b-c)} $
$ \Rightarrow $ $ 2(a-b)(b-c)=-{{(a-c)}^{2}} $
$ \Rightarrow $ $ -{{(a-c)}^{2}}=2(ab-2b^{2}+bc)=2b{ a-2\sqrt{ac}+c } $
$ \Rightarrow $ $ -{{{ (\sqrt{a}-\sqrt{c})(\sqrt{a}+\sqrt{c}) }}^{2}}=2b{{(\sqrt{a}-\sqrt{c})}^{2}} $
Þ $ 2b=-{ a+2\sqrt{ac}+c }=-a-2b-c $ or $ a+4b+c=0 $ .
BETA
