text: # Thread: Geometric Sequences + Irrational Numbers
  
  1. ## Geometric Sequences + Irrational Numbers
  
  Question:
  find the general term of Un which as
  u3=3 and u9=3/8
  
  --> i am looking for the common ratio atm and this is what i did
  u3=u1r^2=3 (1)
  u9=u1r^(-5/8)=3/8 (2)
  
  --> (2)/(1)
  u1r^(-5/8)/ u1r^2 =3/8 / 3
  
  r^(-21/8) = 1/8
  
  r = (-21/8)* root 1/8
  
  But i don't know how to simplify it from there..
  
  but i don't know how to get that
  
  thank you
  
  2. Originally Posted by juliak
  Question:
  find the general term of Un which as
  u3=3 and u9=3/8
  
  --> i am looking for the common ratio atm and this is what i did
  u3=u1r^2=3 (1)
  u9=u1r^(-5/8)=3/8 (2)
  
  --> (2)/(1)
  u1r^(-5/8)/ u1r^2 =3/8 / 3
  
  r^(-21/8) = 1/8
  
  r = (-21/8)* root 1/8
  
  But i don't know how to simplify it from there..
  
  but i don't know how to get that
  
  thank you
  $\displaystyle u_3 = a r^2$ and $\displaystyle u_9 = a r^8$. Therefore:
  
  $\displaystyle \frac{u_9}{u_3} = r^6 = \frac{3/8}{3}\Rightarrow r^6 = \frac{1}{8} \Rightarrow r^6 = \frac{1}{2^3} = ....$
  
  3. u9=a1r^-5/8?
  
  how does u9=a1r^8?
  
  4. Originally Posted by juliak
  u9=a1r^-5/8?
  
  how does u9=a1r^8?
  Didn't you say it's a geometric series.
  
  You're expected to know that the nth term is given by $\displaystyle u_n = a r^{n-1}$.
  
  I have absolutely no idea where you're getting u9=a1r^-5/8 from.
  
  5. oh >_<
  
  but umm i stil don't really get how to simplify
  r^6 = 1/2^3
  why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
  
  6. Originally Posted by juliak
  oh >_<
  
  but umm i stil don't really get how to simplify
  r^6 = 1/2^3
  why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
  You should go back and review all the material (like index laws) that this question clearly implies you should already know.
  
  $\displaystyle r^6 = \frac{1}{2^3} \Rightarrow r = \left( \frac{1}{2^3} \right)^{1/6} = \frac{1}{(2^3)^{1/6}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$.
explanation: The document is a discussion thread about finding the common ratio of a geometric sequence. It contains a math question and a series of responses that provide guidance and explanations to solve the problem. The initial post presents the problem and an attempt at solving it, while subsequent posts clarify and correct the approach.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Thread: Geometric Sequences + Irrational Numbers
  
  1. ## Geometric Sequences + Irrational Numbers
  
  Question:
  find the general term of Un which as
  u3=3 and u9=3/8
  
  --> i am looking for the common ratio atm and this is what i did
  u3=u1r^2=3 (1)
  u9=u1r^(-5/8)=3/8 (2)
  
  --> (2)/(1)
  u1r^(-5/8)/ u1r^2 =3/8 / 3
  
  r^(-21/8) = 1/8
  
  r = (-21/8)* root 1/8
  
  But i don't know how to simplify it from there..
  
  but i don't know how to get that
  
  thank you
  
  2. Originally Posted by juliak
  Question:
  find the general term of Un which as
  u3=3 and u9=3/8
  
  --> i am looking for the common ratio atm and this is what i did
  u3=u1r^2=3 (1)
  u9=u1r^(-5/8)=3/8 (2)
  
  --> (2)/(1)
  u1r^(-5/8)/ u1r^2 =3/8 / 3
  
  r^(-21/8) = 1/8
  
  r = (-21/8)* root 1/8
  
  But i don't know how to simplify it from there..
  
  but i don't know how to get that
  
  thank you
  $\displaystyle u_3 = a r^2$ and $\displaystyle u_9 = a r^8$. Therefore:
  
  $\displaystyle \frac{u_9}{u_3} = r^6 = \frac{3/8}{3}\Rightarrow r^6 = \frac{1}{8} \Rightarrow r^6 = \frac{1}{2^3} = ....$
  
  3. u9=a1r^-5/8?
  
  how does u9=a1r^8?
  
  4. Originally Posted by juliak
  u9=a1r^-5/8?
  
  how does u9=a1r^8?
  Didn't you say it's a geometric series.
  
  You're expected to know that the nth term is given by $\displaystyle u_n = a r^{n-1}$.
  
  I have absolutely no idea where you're getting u9=a1r^-5/8 from.
  
  5. oh >_<
  
  but umm i stil don't really get how to simplify
  r^6 = 1/2^3
  why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
  
  6. Originally Posted by juliak
  oh >_<
  
  but umm i stil don't really get how to simplify
  r^6 = 1/2^3
  why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
  You should go back and review all the material (like index laws) that this question clearly implies you should already know.
  
  $\displaystyle r^6 = \frac{1}{2^3} \Rightarrow r = \left( \frac{1}{2^3} \right)^{1/6} = \frac{1}{(2^3)^{1/6}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread about finding the common ratio of a geometric sequence. It contains a math question and a series of responses that provide guidance and explanations to solve the problem. The initial post presents the problem and an attempt at solving it, while subsequent posts clarify and correct the approach.
  
  Label: question-and-answer