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.
  
  $\displaystyle r^6 = \frac{1}{2^3} \Rightarrow r = \left( \frac{1}{2^3} \right)^{1/6} = \frac{1}{(2^3)^{1/6}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$.