A) A.P.
B) G.P.
C) H.P.
D) none of these
Correct Answer: D
Solution :
[d] a, b, and c are in A.P. Hence, \[2b=a+c\] ...(1) \[\frac{a}{bc}+\frac{2}{b}=\frac{a+2c}{bc}\ne \frac{2}{c}\] \[\Rightarrow \frac{a}{bc},\frac{1}{c},\frac{2}{b}\] are not in A.P. \[\frac{bc}{a}+\frac{b}{2}=\frac{2bc+ab}{2a}\ne c\] Hence, the given numbers are not in H.P. Again, \[\frac{a}{bc}\frac{2}{b}=\frac{2a}{{{b}^{2}}c}\ne \frac{1}{{{c}^{2}}}\] Therefore, the given numbers are not in G.P.
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