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Old November 7th, 2008, 11:25 PM
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Default 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..

The answer said 1/root2
but i don't know how to get that

Help please?
thank you
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  #2  
Old November 7th, 2008, 11:45 PM
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Quote:
Originally Posted by juliak View Post
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..

The answer said 1/root2
but i don't know how to get that

Help please?
thank you
u_3 = a r^2 and u_9 = a r^8. Therefore:

\frac{u_9}{u_3} = r^6 = \frac{3/8}{3}\Rightarrow r^6 = \frac{1}{8} \Rightarrow r^6 = \frac{1}{2^3} = ....
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Old November 7th, 2008, 11:54 PM
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u9=a1r^-5/8?

how does u9=a1r^8?
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Old November 7th, 2008, 11:58 PM
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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 u_n = a r^{n-1}.

I have absolutely no idea where you're getting u9=a1r^-5/8 from.
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Old November 8th, 2008, 12:06 AM
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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?
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Old November 8th, 2008, 12:24 AM
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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.

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}}.
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