Maths-
General
Easy

Question

Let a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma a subscript 4 end subscript and a subscript 5 end subscript be such that a subscript 1 end subscript comma a subscript 2 end subscript and a subscript 3 end subscript are in A.P., a subscript 2 end subscript comma a subscript 3 end subscript and a subscript 4 end subscript are in G.P., and a subscript 3 end subscript comma a subscript 4 end subscript and a subscript 5 end subscript are in H.P. Then log subscript e end subscript invisible function application a subscript 1 end subscript comma log subscript e end subscript invisible function application a subscript 3 end subscript and log subscript e end subscript invisible function application a subscript 5 end subscript are in

  1. G.P.   
  2. A.P.   
  3. H.P.   
  4. None of these  

The correct answer is: A.P.


    We have,
    a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript are in A.P. rightwards double arrow 2 a subscript 2 end subscript equals a subscript 1 end subscript plus a subscript 3 end subscript (1)
    a subscript 2 end subscript comma a subscript 3 end subscript comma a subscript 4 end subscript are in G.P. rightwards double arrow a subscript 3 end subscript superscript 2 end superscript equals a subscript 2 end subscript a subscript 4 end subscript (2)
    a subscript 3 end subscript comma a subscript 4 end subscript comma a subscript 5 end subscript are in H.P. rightwards double arrow a subscript 4 end subscript equals fraction numerator 2 a subscript 3 end subscript a subscript 5 end subscript over denominator a subscript 3 end subscript plus a subscript 5 end subscript end fraction (3)
    Putting a subscript 2 end subscript equals fraction numerator a subscript 1 end subscript plus a subscript 3 end subscript over denominator 2 end fraction and a subscript 4 end subscript equals fraction numerator 2 a subscript 3 end subscript a subscript 5 end subscript over denominator a subscript 3 end subscript plus a subscript 5 end subscript end fraction in (2), we get
    a subscript 3 end subscript superscript 2 end superscript equals fraction numerator a subscript 1 end subscript plus a subscript 3 end subscript over denominator 2 end fraction cross times fraction numerator 2 a subscript 3 end subscript a subscript 5 end subscript over denominator a subscript 3 end subscript plus a subscript 5 end subscript end fraction
    rightwards double arrow a subscript 3 end subscript superscript 2 end superscript equals a subscript 1 end subscript a subscript 5 end subscript
    Hence, a subscript 1 end subscript comma a subscript 3 end subscript, and a subscript 5 end subscript are in G.P. So, log subscript e end subscript invisible function application a subscript 1 end subscript comma log subscript e end subscript invisible function application a subscript 3 end subscript and log subscript e end subscript invisible function application a subscript 5 end subscript are in A.P.

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