Sequence And Series Question 597
Question: If $ 1,\ {\log_{y}}x,\ {\log_{z}}y,\ -15{\log_{x}}z $ are in A.P., then
Options:
A) $ z^{3}=x $
B) $ x={y^{-1}} $
C) $ {z^{-3}}=y $
D) $ x={y^{-1}}=z^{3} $
E) All the above
Show Answer
Answer:
Correct Answer: E
Solution:
Let $ d $ be the common difference then $ {\log_{y}}x=1+d $
$ \Rightarrow $ $ x={y^{1+d}} $ $ {\log_{z}}y=1+2d $
$ \Rightarrow $ $ y={z^{1+2d}} $ and $ -15{\log_{x}}z=1+3d $
$ \Rightarrow $ $ z={x^{-(1+3d)/15}} $
$ \therefore $ $ x={y^{1+d}}={z^{(1+2d)(1+d)}}={x^{-(1+d)(1+2d)(1+3d)/15}} $
$ \Rightarrow $ $ (1+d)(1+2d)(1+3d)=-15 $
$ \Rightarrow $ $ 6d^{3}+11d^{2}+6d+16=0 $
$ \Rightarrow $ $ (d+2)(6d^{2}-d+8)=0 $
$ \Rightarrow $ $ d=-2 $ [Note that $ 6d^{2}-d+8=0 $ has complex roots]
$ \therefore $ $ x={y^{1+d}}={y^{-1}},\ y={z^{1-4}}={z^{-3}} $
$ \therefore $ $ x={{({z^{-3}})}^{-1}}=z^{3} $ . Also $ x={y^{-1}}=z^{3} $ .
BETA
