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  1. #1 Help! 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can anyone solve this question please?

    In an infinite geometrical progression the sum of the first two terms is 6, and every term is four times the sum of all the terms that follow it.
    a)find G.P b)Find the sum to infinite series

    thank you


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  3. #2  
    Forum Ph.D. Leszek Luchowski's Avatar
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    This:

    "every term is four times the sum of all the terms that follow it"

    is a clue to the the ratio of successive terms in the series. Deduce or find the general formula for the sum of a geometric series and you should be able to find the ratio of your series from there.

    Then, use this assertion:

    "the sum of the first two terms is 6"

    to identify the first term.

    Good luck, I know you can do it!


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  4. #3  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Here is how i did, find any possible mistakes.. thank you;



    as is less than 1(its infinite series)

    but here is another clue;

    since

    --which one is bigger??

    =

    =thus clearly the is bigger as its now;



    ----is greater than ----


    common sense tells us that when we divide a number by a number less than 0, we get a larger number. (1-r) is already less than one as r is less than 1, but (1-r)^{2} is even smaller than (1-r) thus when dividing by (1-r^{2}), we get a larger number.

    now drawing a conclusion;
    number

    thus;
    not as the first fraction is a larger number subtracting a smaller number giving a positive integer, and the second fraction is a smaller number subtracted by a larger number thus giving a negative result which cant be.

    two equations;
    --------

    -------



    -----

    thus;






    i would appreciate any advice to my mistakes.
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  5. #4  
    Forum Ph.D. Leszek Luchowski's Avatar
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    I don't have the time to read all your formulas, but this is wrong:

    Quote Originally Posted by Heinsbergrelatz
    common sense tells us that when we divide a number by a number less than 0, we get a larger number.




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  6. #5  
    Forum Ph.D. Leszek Luchowski's Avatar
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    I started reading your formulas and very quickly ran into something I don't understand - I mean, I don't know how you got it:

    Quote Originally Posted by Heinsbergrelatz
    as is less than 1(its infinite series)
    If, for example, , and , then the left side is and the right side is , obviously not equal.
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  7. #6  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    im leaving it as "r" is always less than one(it has to be less than one as it is an infinite series), thus in this question only, when we divide we will always get a larger number.

    infinite series the general formula is im sure you are aware of that, when |r| or the common ration is less than one.

    i meant it as == where |r| is less than one
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  8. #7  
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    you have







    And so,





    I'm pretty sure you can do the rest
    Wise men speak because they have something to say; Fools, because they have to say something.
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  9. #8  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    thnks arcane, but would you say my method is completely wrong??

    also i have some questions on your sigma notation, why cant the i=o, also if you get the i=1 sigma its like which isn't true as the first two terms are . Am i misunderstanding the Sigma notation??
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  10. #9  
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    Quote Originally Posted by Heinsbergrelatz
    thnks arcane, but would you say my method is completely wrong??

    also i have some questions on your sigma notation, why cant the i=o, also if you get the i=1 sigma its like which isn't true as the first two terms are . Am i misunderstanding the Sigma notation??
    If you notice, you arrived at the correct answer, the same as mine, we just got there differently. I found your method a bit convoluted, and hard to follow.

    as to my notation, yes, i can equal 0, but it isn't necessary. I can define my a to one term further back than what you found yours to be. its an arbitrary value for us, really. I can stake it at any point in the entire series, either larger or smaller than the initial value. My a was equal to 25. yours was equal to 5.

    You have a, as demonstrated, proper understanding of the notation.
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  11. #10  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    If you notice, you arrived at the correct answer, the same as mine, we just got there differently. I found your method a bit convoluted, and hard to follow.

    as to my notation, yes, i can equal 0, but it isn't necessary. I can define my a to one term further back than what you found yours to be. its an arbitrary value for us, really. I can stake it at any point in the entire series, either larger or smaller than the initial value. My a was equal to 25. yours was equal to 5.

    You have a, as demonstrated, proper understanding of the notation.
    thank you Arcane, i got it now
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