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A geometric series question

  1. Jun 17, 2005 #1
    I've got a problem here....

    A geometric series has first term 1,the sum of the first 5 terms is twice that of the sum of the 6th to 15th term inclusive. Prove that [tex] r^5= \frac{1}{2} \sqrt {3-1}[/tex]

    What i did was.....

    [tex] 2s_5=s_{15}-s_5 [/tex]

    using the formula for the sum of a GS, i got...

    [tex] 2r^4 -1 =r^{14} -r^4 [/tex]

    and things when downhill from there.

    I've tried expressing it in terms of x^5 because the answer seems to suggest the quadratic formula. But i don't seem to be getting anywhere. Can anyone help?
     
  2. jcsd
  3. Jun 17, 2005 #2

    VietDao29

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    Hmm, it may be me with a wrong calculation, or is it the answer wrong???
    Anyway, I'll give you a hint:
    [tex]A_1 = 1[/tex]
    [tex]A_n = r ^ {n - 1}[/tex]
    So [itex]A_6 = ?[/itex]
    If you assume [itex]A_6[/itex] the first term of another geometric series, with the same r, can you find the sum of the first 10 terms (6th - 15th)?
    Viet dao,
     
    Last edited: Jun 17, 2005
  4. Jun 17, 2005 #3

    StatusX

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    A few things. Is that 3-1? Or, as I like to call it, 2? Also, the formula for the sum of a geometric series, from r^n to r^m is (r^(m+1) - r^n)/(r-1). And finally, didn't you say the sum of first set was twice the sum of the second? Because it looks like you have that backwards in your equation.
     
  5. Jun 17, 2005 #4

    HallsofIvy

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    The first term is one so this series is just 1+ r+ r2+ r3+ ... for some r.
    The sum of the first five terms is [tex]\frac{1-r^6}{1-r}[/tex] and the sum of the "6th to 15th term" is "sum of first 15 terms minus sum of first 5 terms":
    [tex]\frac{1-r^{16}}{1-r}-\frac{1-r^6}{1-r}= \frac{r^6-r^16}{1-r}[/tex].
    We are told that "sum of the first 5 terms is twice that of the sum of the 6th to 15th term" so solve the equation [tex]\frac{1-r^6}{1-r}= \frac{r^6-r^16}{1-r}[/tex]. Solve that for r5.
     
  6. Jun 17, 2005 #5

    StatusX

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    I think those powers should be 5 and 15. The fifth term of the series is r^4. And the actual answer has the 1 outside the radical, but still multiplied by 1/2.
     
  7. Jun 17, 2005 #6

    VietDao29

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    Isn't this the sum of the first 5 terms:
    [tex]S_5 = 1 + r + r^2 + r^3 + r^4 = \frac{r^5 - 1}{r - 1}[/tex]
    Or this:
    [tex]S_5 = 1 + r + r^2 + r^3 + r^4 + r^5= \frac{r^6 - 1}{r - 1}[/tex]?????
    Viet Dao,
     
  8. Jun 18, 2005 #7

    OlderDan

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    I agree with this. Is the negative solution invalid? I don't see why it would be. There is no reason why the series has to converge for infinite n.

    [tex]
    r^5 = \frac{{ \pm \sqrt 3 - 1}}{2} \ \ ??
    [/tex]
     
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