math

in an arithmetic sequence the common difference is equal to 2.the first term is also the first term of a geometric sequence. the sum of the first 3 terms of an arithmetic sequence and the sum of the first 9 terms of an arithmetic sequence form the 2nd and 3rd terms of a geometric sequence respectively determine the first 3 terms of the geometric sequence...How do I go about solving this?

asked by Brainblocked
  1. first term: a
    common difference = 2

    sum of first 3 terms
    = a + a+d + a+2d
    = 3a + 6
    sum of first 9 terms
    = (9/2)(2a + 8(2))
    = 9(a + 8)
    = 9a + 72

    so 3a+6 = ar
    and 9a+72 = ar^2

    square 3a+6 = ar^2
    9a^2 + 36a + 36 = a^2r^2
    r^2= (9a^2 + 36a + 36)/a^2

    from the other equation,
    r^2 = (9a+72)/a

    (9a^2 + 36a + 36)/a^2 = (9a+72)/a
    both sides times a
    (9a^2 + 36a + 36)/a = (9a+72)
    cross-multiply
    9a^2 + 36a + 36 = 9a^2 + 72a
    36a = 36
    a = 1

    in 3a+6 = ar
    3+6 = r
    r = 9

    So the first 3 terms of the GS are 1 , 9, and 81

    check:
    sum of first 3 terms of AS=1 + 3 + 5
    sum of first 9 terms of AS = (9/2)(2 + 16) = 81
    Well, that checked out nicely.

    posted by Reiny

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