# Geometric Series

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?

## 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!

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Dick
Homework Helper
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?

cronxeh
Gold Member
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?

## 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!
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:
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?

cronxeh
Gold Member
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?
Yess