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Math Help - Question on geometric progression (2)

  1. #1
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    Question on geometric progression (2)

    Hello! Please show how to solve this question! I don't know how to solve the second part of the question...

    In a geometric progression, the first term is 12, and the fourth term is (-3/2).

    i)Find the sum of the first n terms of the progression, Sn.

    ii) Find the sum to infinity, S, of the progression [I found this to be 8] and the least value of n for which the magnitude of the difference between Sn and S is less than 0.001.

    Thank you very much!!
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  2. #2
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    Quote Originally Posted by Tangera View Post
    Hello! Please show how to solve this question! I don't know how to solve the second part of the question...

    In a geometric progression, the first term is 12, and the fourth term is (-3/2).

    i)Find the sum of the first n terms of the progression, Sn.

    ii) Find the sum to infinity, S, of the progression [I found this to be 8] and the least value of n for which the magnitude of the difference between Sn and S is less than 0.001.

    Thank you very much!!
    Since this is a geometric progression, our terms will be 12, 12r, 12r^2, 12r^3, ... so the fourth term of the series is 12r^3= -\frac{3}{2}.

    The sum to infinity is \frac{12}{1-r}. So you can find this after solving the previous equation for r.
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    Hello! Thank you for replying! Could you explain how to solve part (ii) of the question too? The part regarding the least value of n....Thank you!
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  4. #4
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    Hello,

    How much do you get for the progression ? Because it's not 8

    S_n=12 \times \frac{1-r^{n+1}}{1-r}

    So S_n-S=\frac{12}{1-r} ((1-r^{n+1})-1)

    \Longleftrightarrow S_n-S=\frac{12}{1-r} (-r^{n+1})

    You're looking for n such as |S_n-S|=\frac{12 |r|^{n+1}}{1-r} < 0.001

    Now plugin r in this expression and solve for n
    Last edited by Moo; April 27th 2008 at 02:46 AM.
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    Quote Originally Posted by Moo View Post
    Hello,

    How much do you get for the progression ? Because it's not 8
    Hello! I found the Sum to infinity to be 8...and that's also the answer that is in the answer key, so it should be correct.
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  6. #6
    Moo
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    Oh sorry, I was talking about r, not S !
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    Quote Originally Posted by Moo View Post
    Oh sorry, I was talking about r, not S !
    oh...I found r = -1/2 and S = 8...
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  8. #8
    Moo
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    Yes, it's ok !

    And what did you find for n ? Do you manage to do it ?
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    noooo....the expression I got for Sn was 8[1 - (-1/2)^n] and then i couldn't solve the modulus...
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  10. #10
    Moo
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    S_n=8[1 - (-1/2)^n]
    It's ok, it's just n+1 instead of n because there are n+1 terms in S_n

    S_n=8-8(-1)^{n-1}(.5)^{n+1}

    You have to take the absolute value of the difference, because S is constant and the sign always change for the terms.

    |S_n-S|=|S_n-8|=|{\color{red}-}8 {\color{red}(-1)^{n-1}}(.5)^{n+1}|=8(.5)^{n+1}
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    But I have a (-1/2)^n so it is impossible to solve the modulus...?
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  12. #12
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    Take the absolute value of -1/2, that is to say 1/2. You want to care about the magnitude, that is to say the absolute difference between the two.
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    HellO! [sorry this is reeaaaally late.] Thank you for replying! I have solved the question.
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