A) \[{{z}^{3}}=x\]
B) \[x={{y}^{-1}}\]
C) \[{{z}^{-3}}=y\]
D) \[x={{y}^{-1}}={{z}^{3}}\]
E) All the above
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 \[-15{{\log }_{x}}z=1+3d\]\[\Rightarrow \]\[z={{x}^{-(1+3d)/15}}\] \[\therefore \]\[x={{y}^{1+d}}={{z}^{(1+2d)(1+d)}}={{x}^{-(1+d)(1+2d)(1+3d)/15}}\] \[\Rightarrow \]\[(1+d)(1+2d)(1+3d)=-15\] \[\Rightarrow \]\[6{{d}^{3}}+11{{d}^{2}}+6d+16=0\] \[\Rightarrow \]\[(d+2)(6{{d}^{2}}-d+8)=0\]\[\Rightarrow \]\[d=-2\][Note that \[6{{d}^{2}}-d+8=0\] has complex roots] \[\therefore \] \[x={{y}^{1+d}}={{y}^{-1}},\ y={{z}^{1-4}}={{z}^{-3}}\] \[\therefore \]\[x={{({{z}^{-3}})}^{-1}}={{z}^{3}}\]. Also\[x={{y}^{-1}}={{z}^{3}}\].
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