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Math Help - another arithmetic series problem

  1. #1
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    Smile another arithmetic series problem

    an artihmetic series has a first term of 3 and a common difference of 2

    determine the number of terms needed for te sum to exceed 1000

    I think I'm missing something blindingly obvious but i can't seem to finish it, but at the moment i'm this far...


    1000< n/2(2n+4)

    now I'm stuck....

    any help is greatly recieved
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by pop_91 View Post
    an artihmetic series has a first term of 3 and a common difference of 2

    determine the number of terms needed for te sum to exceed 1000

    I think I'm missing something blindingly obvious but i can't seem to finish it, but at the moment i'm this far...


    1000< n/2(2n+4)

    now I'm stuck....

    any help is greatly recieved
    Perfect so far


    Well now :
    \frac n2 (2n+4)=n(n+2)

    So you want to solve n(n+2)>1000
    n^2+2n>1000

    n^2+2n+1-1>1000

    (n+1)^2-1>1000

    (n+1)^2>1001

    In fact, if you want to find the minimum value of n, solve (n+1)^2=1001 and then round it up by superior value.

    Does it look clear ?
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  3. #3
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    Smile

    yes i think i understand just one question...

    do you not have to divide 1000 by 2 because you have to do the same to both sides.... i'm not sure
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  4. #4
    Moo
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    Quote Originally Posted by pop_91 View Post
    yes i think i understand just one question...

    do you not have to divide 1000 by 2 because you have to do the same to both sides.... i'm not sure
    Oh !

    2n+4=2(n+2)
    so \frac n2 (2n+4)=\frac{n \cdot {\color{red}2} (n+2)}{\color{red}2}=n(n+2)
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  5. #5
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    oooh! thanks i get it now x
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