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

  • question_answer
    If the \[{{(m+1)}^{th}},\ {{(n+1)}^{th}}\] and \[{{(r+1)}^{th}}\] terms of an A.P. are in G.P. and \[m,\ n,\ r\] are in H.P., then the value of the ratio of the common difference to the first term of the A.P. is    [MNR 1989; Roorkee 1994]

    A) \[-\frac{2}{n}\]

    B) \[\frac{2}{n}\]

    C) \[-\frac{n}{2}\]

    D) \[\frac{n}{2}\]

    Correct Answer: A

    Solution :

    Let \[a\] be the first term and d be the common difference of the given A.P. Then as given the \[{{(m+1)}^{th}}\], \[{{(n+1)}^{th}}\]and \[{{(r+1)}^{th}}\] terms are in G.P. \[\Rightarrow \]\[a+md,\ a+nd,\ a+rd\]  are in G.P. \[\Rightarrow \]\[{{(a+nd)}^{2}}=(a+md)(a+rd)\] \[\Rightarrow \]\[a(2n-m-r)=d(mr-{{n}^{2}})\] or \[\frac{d}{a}=\frac{2n-(m+r)}{mr-{{n}^{2}}}\] ?..(i) Next, \[m,\ n,\ r\] in H.P. \[\Rightarrow n=\frac{2mr}{m+r}\] ?..(ii) From (i) and (ii) \[\frac{d}{a}=\frac{2n-(m+r)}{mr-{{n}^{2}}}=\frac{2}{n}\left( \frac{2n-(m+r)}{(m+r)-2n} \right)=-\frac{2}{n}\].


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