Geometric Series question

[ref91]

The answer is 13, any ideas on how to get there?

 

[Deleted member 4993]

View attachment 22676

The answer is 13, any ideas on how to get there?

Use logarithm.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[ref91]

I assumed this is a geometric series question and calculated that r = 1/3. U1 also = 1/3. So U_n=U₁ x r^n-1

so 1/100000= 1/3 x (1/3)^n-1

1=100000/3 x (1/3)^n-1
3^n-1= 100000/3

And then I got stuck

 

[Steven G]

I have two points to make. The first one is that 13 is not in this sequence and 13 is NOT less than 10^-6 which is less than 1.

Possibly you mean the 13th term?? Well what is the 13th terms and is it less than 10^(-6)? Is the 12th term less than 10^(-6)?

You have to try something!

 

[Deleted member 4993]

I assumed this is a geometric series question and calculated that r = 1/3. U1 also = 1/3. So U_n=U₁ x r^n-1

so 1/100000= 1/3 x (1/3)^n-1

1=100000/3 x (1/3)^n-1
3^n-1= 100000/3

And then I got stuck

Did you understand the first line of response #2.

 

[Steven G]

I assumed this is a geometric series question and calculated that r = 1/3. U1 also = 1/3. So U_n=U₁ x r^n-1

so 1/100000= 1/3 x (1/3)^n-1

1=100000/3 x (1/3)^n-1
3^n-1= 100000/3

And then I got stuck

Why are you talking about n-1. This makes things more complicated.

The 1st term is 1/3¹
The 2nd term is 1/3²
The 3rd term is 1/3³
The 4th term is 1/3⁴
Do you see a simple pattern here??
The nth term is 1/3ⁿ

Just solve for the smallest n such that 1/3ⁿ<10^-6=1/10⁶ which is the same as 3ⁿ>10⁶.


If you want to continue your way then you should now multiply both sides by 3

 

[Steven G]

using logs work but you can do this by trial and error.

Pick n=10. Is 3^10 >10^6? If not then trian an even larger value for n. If yes, then see if n=9 satisfies the inequality.


To get a handle on what the answer will be close to note that 3^2=9 is close to 10 but less than 10. So 3^12 = (3^2)^6 = 9^6<10^6 so the power of 3 must be more than 12 but since 9 is so close to 10 it will not be much larger than 12 (ie I would not try n=125 after seeing that n=12 did not work)

 

[ref91]

Why are you talking about n-1. This makes things more complicated.

The 1st term is 1/3¹
The 2nd term is 1/3²
The 3rd term is 1/3³
The 4th term is 1/3⁴
Do you see a simple pattern here??
The nth term is 1/3ⁿ

Just solve for the smallest n such that 1/3ⁿ<10^-6=1/10⁶ which is the same as 3ⁿ>10⁶.


If you want to continue your way then you should now multiply both sides by 3

Thank you for explaining it so clearly
So log₃ 10⁶=12.57... so answer is 13?

 

[ref91]

Did you understand the first line of response #2.

Apologies