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Geometric progression

  1. Apr 7, 2005 #1
    Hi there,
    Can anybody help please.

    How can i find the first term in a geometric progression if i know that the 4th term = 256 and the 8th term = 65536?
     
  2. jcsd
  3. Apr 7, 2005 #2
    Those are both powers of 2.
     
  4. Apr 7, 2005 #3
    Sorry, perhaps that wasn't a good example, in general how do i find the 1st term (say a) in a geometric progression if i know 2 other other terms (say b and c)
     
  5. Apr 7, 2005 #4

    dextercioby

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    Use logarithmation.If 65536 is the 8-th term,then the first is [itex] \log_{8} 65536 [/itex]...(the first is just the ratio.If it's not,then u need two terms...)

    Daniel.
     
  6. Apr 7, 2005 #5
    It turns out the answer is -4

    There must be a formula to work this out without using logarithmation
     
  7. Apr 7, 2005 #6
    Actually, it's [itex]\pm 4[/itex]. See why?
     
  8. Apr 7, 2005 #7

    matt grime

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    What is the definition of the n'th term of a GP with initial term a and ratio r?

    ar^{n-1}, right?


    so given two terms you've two unknowns and fortunately you can solve for them, though at some point you will need to take some roots.
     
  9. Apr 7, 2005 #8

    HallsofIvy

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    The general term of a geometric sequence is arn-1 where a is the first value (n= 1) and r is the "common ratio". If you know "the i th term is x" then you know ari-1= x. If you know "the jth term is y" then you know arj-1= y. Divide the second equation by the first and the "a"s cancel: rj-1/ri-1= rj-i= y/x. Now it's not really necessary to use logarithms to find r- just use the "j-i" root. Once you know r, you can solve either of the original equations for a.

    In the example you started with, 4th term = 256 and the 8th term = 65536,
    we know ar3= 256 and ar7= 65536 so, dividing the second equation by the first, r7/r3= r4= 65536/256= 256.
    Since r= 256, a(2563)= 256 = 1/2562= 1/65536.
     
    Last edited: Apr 7, 2005
  10. Apr 7, 2005 #9

    dextercioby

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    Halls,[itex] r^{4}=256 [/itex] has 4 solutions...[itex] \left\{\pm 4,\pm 4i\right\} [/itex]...

    Daniel.
     
  11. Apr 7, 2005 #10
    you mean [itex]r^4 = 256 \Longrightarrow r = \pm 4[/itex] (over [itex]\mathbb{R}[/itex]) so

    [tex]a(\pm 4)^3 = 256 \Longrightarrow a =\biggr \{ \begin{array}{cc} 4 & \mbox{if} \ r=4 \\ -4 & \mbox{if} \ r = -4\end{array}[/tex]

    or just [itex]a = r = \pm 4[/itex].
     
  12. Apr 7, 2005 #11

    HallsofIvy

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    Oops. My math is fine- but my arithmetic is terrible!
     
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