A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
Let \[{{x}^{a}}={{x}^{b/2}}{{z}^{b/2}}={{z}^{c}}=\lambda \] \[\Rightarrow \]\[x={{\lambda }^{1/a}},\ z={{\lambda }^{1/c}},\ xz={{\lambda }^{2/b}}\] \[\Rightarrow \]\[{{\lambda }^{(1/a)+(1/c)}}={{\lambda }^{2/b}}\]\[\Rightarrow \]\[\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\] \[\Rightarrow \]\[a,\ b,\ c\] are in H.P.
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