A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
Let \[{{a}^{x}}={{b}^{y}}={{c}^{z}}={{d}^{u}}=k\] (say) Then \[a={{k}^{1/x}},\ b={{k}^{1/y}},\ c={{k}^{1/z}},\ d={{k}^{1/u}}\] Since \[a,\ b,\ c\] are in G.P., therefore \[{{b}^{2}}=ac\Rightarrow {{k}^{2/y}}={{k}^{1/x}}.\,\,{{k}^{1/z}}={{k}^{1/x+1/z}}\] \[\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}\Rightarrow \frac{1}{x},\ \frac{1}{y},\ \frac{1}{z}\] are in A.P. \[\Rightarrow \]\[x,\ y,\ z\] are in H.P. Similarly it can be shown that \[y,\ z,\ u\] are also in H.P. \[\therefore \]\[x,\ y,\ z\] and \[u\] are in H.P.
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