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

  • question_answer
    When \[\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-d}=0\] and \[b\ne a\ne c\], then \[a,\ b,\ c\] are [MP PET 2004]

    A) In H.P.

    B) In G.P.

    C) In A.P.

    D) None of these

    Correct Answer: A

    Solution :

    We have \[\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-b}=0\] \[\frac{1}{a}+\frac{1}{c-b}=\frac{1}{b-a}-\frac{1}{c}\] \[\Rightarrow \] \[\frac{c-b+a}{a(c-b)}=\frac{c-b+a}{(b-a)c}\]\[\Rightarrow \]\[ac-ab=bc-ac\] \[\Rightarrow \]\[2ac=ab+bc\]\[\Rightarrow \]\[\frac{2ac}{a+c}=b\]i.e., \[a,b,c\] are in H.P.


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