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suppose an arithmetic sequence and a geometric sequence with common ration r have the same first two terms. show that the third term of the geometric series is r^2/(2r-1) tomes the third term of the arithmetic sequence

  • math -

    the first 3 terms of the AS are a, a+d, a+2d
    the first 3 terms of the GS are a, ar, ar^2

    you said the second terms are equal
    then a+d = ar
    d = ar-a

    so third term of GS/third term of AS
    = ar^2/(a+2d)
    = ar^2(a + 2(ar-a)
    = ar^2/(a + 2ar - 2a)
    = ar^2/(2ar - a) now divide top and bottom by a
    = r^2/(2r-1)

  • math -

    swaggot swag

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