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Thread: Geometric Progression #2

  1. #1
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    Geometric Progression #2

    Thanks a lot mrfantastic!!! =)
    But I got a similiar question which I can't solve:

    The sum of the first 20 terms of a geometric series is 10 and the sum of the first 30 terms is 91, find the sum of the first 10 terms.

    I divided S20 by S10 to eliminate 'a' so as to find 'r':
    (r^30 -1) / (r^20 -1) = 91/10

    Got to here but I am not sure how to continue. If I expand I will end up with this equation:
    10r^30 - 91r^20 + 81 = 0
    and the graph of this equation looks kind of weird.

    But if I express in this form: 10(r^10)^3 - 91(r^10)^2 +81 = 0
    I get r^10 = 9, 1, -9/10

    Is there a better way to simplify so I can end up with a simple EQN like the previous question here: http://www.mathhelpforum.com/math-he...tml#post661852?
    Last edited by mr fantastic; Jun 22nd 2011 at 10:14 PM.
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  2. #2
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    Re: Geometric Progression #2

    Quote Originally Posted by Blizzardy View Post
    Thanks a lot mrfantastic!!! =)
    But I got a similiar question which I can't solve:

    The sum of the first 20 terms of a geometric series is 10 and the sum of the first 30 terms is 91, find the sum of the first 10 terms.

    I divided S20 by S10 to eliminate 'a' so as to find 'r':
    (r^30 -1) / (r^20 -1) = 91/10

    Got to here but I am not sure how to continue. If I expand I will end up with this equation:
    10r^30 - 91r^20 + 81 = 0
    and the graph of this equation looks kind of weird.

    But if I express in this form: 10(r^10)^3 - 91(r^10)^2 +81 = 0
    I get r^10 = 9, 1, -9/10

    Is there a better way to simplify so I can end up with a simple EQN like the previous question here: http://www.mathhelpforum.com/math-he...tml#post661852?
    Dear Blizzardy,

    $\displaystyle \frac{a(r^{20}-1)}{r-1}=10$

    $\displaystyle \frac{a(r^{10}-1)(r^{10}+1)}{r-1}=10$

    $\displaystyle S_{10}=\frac{a(r^{10}-1)}{r-1}=\frac{10}{r^{10}+1}$

    You have obtained values for $\displaystyle r^{10}$. Hence there are three possible values that $\displaystyle S_{10}$ could take depending on $\displaystyle r^{10}$. Hope you can continue.
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  3. #3
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    Re: Geometric Progression #2

    Quote Originally Posted by Blizzardy View Post
    Thanks a lot mrfantastic!!! =)
    But I got a similiar question which I can't solve:

    The sum of the first 20 terms of a geometric series is 10 and the sum of the first 30 terms is 91, find the sum of the first 10 terms.

    I divided S20 by S10 to eliminate 'a' so as to find 'r':
    (r^30 -1) / (r^20 -1) = 91/10

    Got to here but I am not sure how to continue. If I expand I will end up with this equation:
    10r^30 - 91r^20 + 81 = 0
    and the graph of this equation looks kind of weird.

    But if I express in this form: 10(r^10)^3 - 91(r^10)^2 +81 = 0
    I get r^10 = 9, 1, -9/10

    Is there a better way to simplify so I can end up with a simple EQN like the previous question here: http://www.mathhelpforum.com/math-he...tml#post661852?
    I think that the way the problem was presented,
    you could try to find S10 from S30 and S20,
    since the difference between these sums is 10 terms.

    $\displaystyle S_{30}=a+ar+ar^2+ar^3+.......+ar^{29}$

    $\displaystyle S_{20}=a+ar+ar^2+ar^3+....+ar^{19}$

    $\displaystyle S_{10}=a+ar+ar^2+ar^3+....+ar^9$

    We can write 2 equations from the above

    $\displaystyle S_{30}-S_{20}=ar^{20}+ar^{21}+....+ar^{29}=r^{20}\left(a+ ar+ar^2+...+ar^9}\right)=r^{20}S_{10}$

    Hence

    $\displaystyle 91-10=81=r^{20}S_{10}$

    Also

    $\displaystyle S_{20}=S_{10}+ar^{10}+ar^{11}+....+ar^{19}=S_{10}+ r^{10}S_{10}$

    $\displaystyle \Rightarrow\ S_{10}\left(1+r^{10}\right)=10$

    Therefore, we have the equation

    $\displaystyle S_{10}=\frac{10}{1+r^{10}}=\frac{81}{r^{20}}$

    This leads to a quadratic equation

    $\displaystyle 10r^{20}=81\left(1+r^{10}\right)$

    $\displaystyle 10\left(r^{10}\right)^2-81r^{10}-81=0$

    $\displaystyle 10x^2-81x-81=0\Rightarrow\ (10x+9)(x-9)=0$

    Since $\displaystyle r^{10}$ is an even power and hence positive, the negative solution for x is ruled out.

    $\displaystyle r^{10}=9$

    Finally

    $\displaystyle S_{10}=\frac{81}{\left(r^{10}\right)^2}=1$
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  4. #4
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    Re: Geometric Progression #2

    Quote Originally Posted by Blizzardy View Post
    Thanks a lot mrfantastic!!! =)
    But I got a similiar question which I can't solve:

    The sum of the first 20 terms of a geometric series is 10 and the sum of the first 30 terms is 91, find the sum of the first 10 terms.

    I divided S20 by S10 to eliminate 'a' so as to find 'r':
    (r^30 -1) / (r^20 -1) = 91/10

    Got to here but I am not sure how to continue. If I expand I will end up with this equation:
    10r^30 - 91r^20 + 81 = 0
    and the graph of this equation looks kind of weird.

    But if I express in this form: 10(r^10)^3 - 91(r^10)^2 +81 = 0
    I get r^10 = 9, 1, -9/10

    Is there a better way to simplify so I can end up with a simple EQN like the previous question here: http://www.mathhelpforum.com/math-he...tml#post661852?
    After seeing Archie Meade's post I found a mistake in my previous post. You cannot use all the three values you have obtained as I have mistakenly stated.

    Case 1: Let, $\displaystyle r^{10}=1\Rightarrow{r=\pm{1}}$

    When r=1;

    $\displaystyle S_{20}=a+ar+ar^2+ar^3+....+ar^{19}=20a=10 \Rightarrow{a=0.5}$

    $\displaystyle S_{30}=a+ar+ar^2+ar^3+.......+ar^{29}=30a=91 \Rightarrow{a=\frac{91}{31}}$

    But 'a' must have a unique value, so we cannot take r=1.

    When r=-1;

    $\displaystyle S_{30}=S_{20}=0$

    But we know that, $\displaystyle S_{30}\mbox{ and }S_{20}$ are not equal to zero. Hence we cannot take r=-1 either.

    So $\displaystyle r^{10}=1$ cannot be taken.

    Case 2: Let, $\displaystyle r^{10}=-\frac{9}{10}$

    In this case r would be a complex value and obviously the sums $\displaystyle S_{30}\mbox{ and }S_{20}$ would also be complex values which is again a contradiction.

    Therefore we cannot take $\displaystyle r^{10}=-\frac{9}{10}$

    The only solution that could be used is, $\displaystyle r^{10}=9$
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