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Math Help - geometric series problem

  1. #1
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    geometric series problem

    In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

    If you take the growth in population made at the same rate , find :

    a) how much is the population grow of the city in the time period 1990-2000 ?

    b) how much became the population number this year(2009) ?

    c) when will the city have 1400000?



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  2. #2
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    Quote Originally Posted by beq!x View Post
    In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

    If you take the growth in population made at the same rate , find :
    What sort of growth are you supposed to assume? Linear? Exponential? Something else?

    When you reply, please include a clear listing of your steps and reasoning so far. Thank you!
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  3. #3
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    Quote Originally Posted by beq!x View Post
    In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

    If you take the growth in population made at the same rate , find :

    a) how much is the population grow of the city in the time period 1990-2000 ?

    b) how much became the population number this year(2009) ?

    c) when will the city have 1400000?



    If the title is assumed to be correct then you can use the general formula for a geometric sequence

    Using the sum of a geometric sequence:

    S_n = \frac{a(r^n-1)}{r-1}

    If you set 1990 as n=0 then in 2000: n = 2000-1990. Setting 1990 as 0 will also give you a value of a from which the common ratio r can be found
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  4. #4
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    this is the original problem i didn't change anything.
    i think at first we have to find q.

    at c) is it after 27 years ? am i right ?
    Last edited by mr fantastic; April 19th 2009 at 06:33 AM. Reason: Merged posts
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