A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
\[\frac{a+b}{1-ab},\ b,\ \frac{b+c}{1-bc}\] are in A.P. \[\Rightarrow \]\[b-\frac{a+b}{1-ab}=\frac{b+c}{1-bc}-b\] \[\Rightarrow \] \[-\frac{a({{b}^{2}}+1)}{1-ab}=\frac{c({{b}^{2}}+1)}{1-bc}\]\[\Rightarrow \]\[-\left( \frac{1-ab}{a} \right)=\frac{1-bc}{c}\] \[\Rightarrow \]\[-\frac{1}{a}+b=\frac{1}{c}-b\]\[\Rightarrow \]\[2b=\frac{1}{a}+\frac{1}{c}\] \[\Rightarrow \] \[a,\ \frac{1}{b},\ c\] are in H.P.
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