A) \[x=y=z\]
B) \[x=y=-z\]
C) \[x=1,y=2,z=3\]
D) \[x=2,y=4,z=6\]
E) \[x=2y=3z\]
Correct Answer: A
Solution :
\[\because \]\[x,\text{ }y,\text{ }z\] are in AP \[\therefore \] \[y=\frac{x+z}{2}\] and\[{{\tan }^{-1}}x,{{\tan }^{-1}}y\]and\[{{\tan }^{-1}}z\]are also in AP. \[2{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}z\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{2y}{1-{{y}^{2}}} \right)={{\tan }^{-1}}\left( \frac{x+z}{1-xz} \right)\] \[\Rightarrow \] \[\frac{2y}{1-{{y}^{2}}}=\frac{2y}{1-xz}\] \[\Rightarrow \] \[{{y}^{2}}=xz\] \[\Rightarrow \] \[x=y=z\]
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