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JEE Main & Advanced Mathematics Sequence & Series Question Bank Relation between AP., GP. and HP.

  • question_answer
    \[x+y+z=15\] if \[9,\ x,\ y,\ z,\ a\] are in A.P.; while \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}\] if \[9,\ x,\ y,\ z,\ a\] are in H.P., then the value of \[a\] will be [IIT 1978]

    A) 1

    B) 2

    C) 3

    D) 9

    Correct Answer: A

    Solution :

    \[x+y+z=15\], if \[x={{({{z}^{-3}})}^{-1}}={{z}^{3}}\] are in A.P. Sum \[=9+15+a=\frac{5}{2}(9+a)\]\[\Rightarrow \]\[24+a=\frac{5}{2}(9+a)\] \[\Rightarrow \]\[48+2a=45+5a\]\[\Rightarrow \]\[3a=3\]\[\Rightarrow \]\[a=1\] ?..(i) and \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}\],  if \[9,\ x,\ y,\ z,\ a\]are in H.P. Sum =\[\frac{1}{9}+\frac{5}{3}+\frac{1}{a}=\frac{5}{2}\left[ \frac{1}{9}+\frac{1}{a} \right]\]\[\Rightarrow \]\[a=1\].


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