A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: C
Solution :
\[\because a,\,\,b,\,\,c\] are in G.P. Þ \[{{b}^{2}}=ac\] ?..(i) Let \[{{a}^{x}}={{b}^{y}}={{c}^{z}}=k\] Þ \[a={{k}^{1/x}},b={{k}^{1/y}},c={{k}^{1/z}}\] Putting these values in (i), \[{{k}^{2/y}}={{k}^{1/x}}.{{k}^{1/z}}\] \[={{k}^{\frac{1}{x}+\frac{1}{z}}}\] i.e., \[\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\] \[\therefore \frac{1}{x},\frac{1}{y},\frac{1}{z}\] are in A.P. or \[x,y,z\] are in H.P.
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