Find k for Geometric Sequence b1=1000, bn=(2/3)bn-1: 0.001

[kazuchan]

b₁,b₂,b₃,...​

In the geometric sequence above, b₁=1000 and b_n=(2/3)b_n-1 for all n$$\geq$$2. What is the least value of k for which b_k<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!

 

[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?

 

[cronxeh]

b₁,b₂,b₃,...​

In the geometric sequence above, b₁=1000 and b_n=(2/3)b_n-1 for all n$$\geq$$2. What is the least value of k for which b_k<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 b_k 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, ... b_k=(2/3)*b_k-1

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

1500*(2/3)^x=1/1000. Find x, and your answer would be for k>x

Edit: I see Dick beat me to it

 

[kazuchan]

Thank you both for your help!

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

 

[cronxeh]

Thank you both for your help!

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

Yess