precal

what is the 7th geometric sequence when a10 is 9 and a13 is -72

asked by gail
  1. 9 9r 9r^2 9r^3

    -72 = 9 r^3
    r^3 = -72/9

    now work back

    a9 = 9/r
    a8 = 9/r^2
    a7 = 9/r^3

    interesting

    a7 = 9/r^3 = (9)(-9/72) = -81/72

    posted by Damon
  2. or ,

    using the standard a to be first term, r the common ratio

    ar^12 = -72
    ar^9 =9
    divide them
    r^3 = -8
    r = -2 , then from ar^9 = 9
    a = 9/(-2)^9 = -9/512

    term(7) = ar^6
    = (-9/512)(-2^6) = -576/512 = -9/8

    posted by Reiny

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