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An arithmetic series has a first term of -4 and a common difference of 1. A geometric series has a first term of 8 and a common ratio of 0.5. After how many terms does the sum of the arithmetic series exceed the sum of the geometric series?

  • math studies -

    anyone????

  • math studies -

    please

  • math studies -

    can anyone help?

  • math studies -

    for the AS
    Sn = n/2(-8 + n-1)
    = n/2(n-9)

    for a GS
    Sn = 8(1-.5^n)/(1-.5)
    so n/2(n-9) > 8(1-.5^n)/(1-.5)
    n/2(n-9) > 16(1-.5^n)
    n^2 - 9n > 32 - 32(.5^n)

    32(.5)^n + n^2 - 9n - 32 > 0

    set it equal to zero and attempt to solve it.

    This would be a very nasty equation to solve, except you know that n has to be a whole number, so do some trial and error calculations.

    eg. n = 2 we get -38 > 0 which is false
    if n = 5 we get -44 > 0 which is false
    if n = 20 we get 188 > 0 which is true

    So somewhere between n=5 and n=20 there should be a solution

    (n=12 ---> 4.0078 > 0 mmmmhhh?)

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