A) \[x,\text{ }y,\text{ }z\]are in GP
B) \[x,\text{ }y,\text{ }z\]are in AP
C) \[\frac{1}{x},\frac{1}{y},\frac{1}{z}\] tare in GP
D) \[\frac{1}{x},\frac{1}{y},\frac{1}{z}\] are in AP
Correct Answer: D
Solution :
Given, \[{{4}^{x}}={{16}^{y}}={{64}^{z}}\] \[\Rightarrow \] \[{{2}^{2x}}={{2}^{4y}}={{2}^{6z}}\] \[\Rightarrow \] \[2x=4y=6z\] \[\Rightarrow \] \[x=2y=3z\] \[\Rightarrow \] \[\frac{x}{6}=\frac{y}{3}=\frac{z}{2}=k\] \[\Rightarrow \] \[x=6k,\,y=3k,\,z=2k\] Let \[\frac{1}{x},\frac{1}{y},\frac{1}{z}\] are in AP. \[\Rightarrow \] \[\frac{1}{6k},\frac{1}{3k},\frac{1}{2k}\] are in AP. \[\Rightarrow \] \[2.\frac{1}{3k}=\frac{1}{6k}+\frac{1}{2k}=\frac{4}{6k}\] \[\Rightarrow \] \[\frac{2}{3k}=\frac{2}{3k}\]
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