# Geometric Series

1. Sep 6, 2010

### kazuchan

b1,b2,b3,......​
In the geometric sequence above, b1=1000 and bn=(2/3)bn-1 for all n$$\geq$$2. What is the least value of k for which bk<0.001?

3. The attempt at a solution
What I did first was I found what b0 is since we are given b1 and that is 1500. But I do not understand where the k is coming from and what needs to be found. Can someone please help me?
Thank you!

2. Sep 6, 2010

### Dick

There's no need to find b0. b1=1000. b2=1000*(2/3). That's bk for k=1. b3=1000*(2/3)*(2/3), right? That's bk for k=3. None of those is less that 0.001. But they are getting smaller. How large does k have to be to make bk<0.001?

3. Sep 6, 2010

### cronxeh

Its asking you for which bk is the sequence starting to be less than 0.001

According to you.. b0 = 1500, b1= 1000, b2=(2/3)*1000, b3=(2/3)*b2, b4=(2/3)*b3, ... bk=(2/3)*bk-1

0.001 is 1/(1000). So how many times would you have to multiply 1500*(2/3)^x to equal 1/1000 ?

Edit: I see Dick beat me to it

Last edited: Sep 6, 2010
4. Sep 6, 2010

### kazuchan

Thank you both for your help!

Now I understand what the problem is asking. So if I continue to plug in each bk i get and try to find the one where bk<.001 the answer would be 36?

5. Sep 6, 2010

Yess