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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    Suppose a, b, c are in AP and | a|, | b|, | c| < 1. If \[x=1+a+{{a}^{2}}+....\,to\,\infty ,\] \[y=1+b+{{b}^{2}}+....\,\text{to}\,\infty ,\] \[z=-1+c+{{c}^{2}}+...\,\text{to}\,\infty \] then \[x,y,z\]are in

    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.


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