A) AP
B) GP
C) HP
D) none of these
Correct Answer: C
Solution :
Given a, b, c are in AP and \[|a|,|b|,|c|\,<\,1\] and \[x=1+a+{{a}^{2}}+...to\infty \] \[y=1+b+{{b}^{2}}+...to\,\infty \] \[z=1+c+{{c}^{2}}+...\,\text{to}\,\infty \] Then, \[x=\frac{1}{1-a},y=\frac{1}{1-b},z=\frac{1}{1-c}\] \[\Rightarrow \] \[a=1-\frac{1}{x},b=1-\frac{1}{y},c=1-\frac{1}{z}\] \[\Rightarrow \] \[\frac{1}{x}=1-a,\frac{1}{y}=1-b,\frac{1}{z}=1-c\] But given a, b, c are in AP \[\therefore \] \[1-a,1-b,1-c\]are in HP. \[\Rightarrow \]\[\frac{1}{x},\frac{1}{y},\frac{1}{z}\]are also in AP \[\Rightarrow \]\[x,y,z\]are in HP.You need to login to perform this action.
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