Geometric Sequences + Irrational Numbers

**juliak** · Nov 7th 2008, 10:25 PM

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

**mr fantastic** · Nov 7th 2008, 10:45 PM

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..

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:

$\frac{u_9}{u_3} = r^6 = \frac{3/8}{3}\Rightarrow r^6 = \frac{1}{8} \Rightarrow r^6 = \frac{1}{2^3} = ....$

**juliak** · Nov 7th 2008, 10:54 PM

u9=a1r^-5/8?

how does u9=a1r^8?

**mr fantastic** · Nov 7th 2008, 10:58 PM

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 $u_n = a r^{n-1}$ .

I have absolutely no idea where you're getting u9=a1r^-5/8 from.

**juliak** · Nov 7th 2008, 11:06 PM

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?

**mr fantastic** · Nov 7th 2008, 11:24 PM

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.

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

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