A) AP
B) GP
C) HP
D) None of these
Correct Answer: A
Solution :
Since, a, b, c are in AP. \[\therefore \] \[2b=a+c\] If\[\frac{1}{bc},\frac{1}{c},\frac{1}{b}\]are in AP, then \[2\left( \frac{1}{c} \right)=\frac{a}{bc}+\frac{1}{b}\] \[\Rightarrow \] \[\frac{2}{c}=\frac{a+c}{bc}\] \[\Rightarrow \] \[\frac{2bc}{c}=a+c\] \[\Rightarrow \] \[2b=a+c\] Hence,\[\frac{a}{bc},\frac{1}{c},\frac{1}{b}\]are in AP.
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