Any chance I can borrow of cup of geometric sequences from some kind (and wise) souls on the forum?

Hired as an English teacher, I frequently help students with math concerns. In particular, is there an online resource that can explain why the reciprocal of the exponent (n-1) is (1-n)?

Is the above even true? I'm looking at an equation that essentially reads (1/3)^n-1 = (3)^1-n.

thanks, Todd's about-to-be (water) logged keyboard

I have no idea. I just clicked your thread to see which keyboard you are using.

Quote
Todd's keyboard
Any chance I can borrow of cup of geometric sequences from some kind (and wise) souls on the forum?

Hired as an English teacher, I frequently help students with math concerns. In particular, is there an online resource that can explain why the reciprocal of the exponent (n-1) is (1-n)?

Is the above even true? I'm looking at an equation that essentially reads (1/3)^n-1 = (3)^1-n.

thanks, Todd's about-to-be (water) logged keyboard

Your example does not work, just plug in 5 for n and it fails to give the same answer for both sides. So something or another is out of place there. However I can't help you with the equation as I had my brain fried by Physics homework last night and am still in rehab.

Lew (and others),

Yes, I think you are correct. I should not have written that with an equal sign. Rather, those terms are reciprocals of each other.

thanks, Todd's multi-tasking-therfore-multi-mistaking board



Edited 1 time(s). Last edit at 09/28/2022 04:17PM by Todd's keyboard.

1/3 = 3^-1

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

Tominator,

Thanks. That makes sense. I'm embarrassed that I didn't see it before.

Perhaps to make it more clear to students, I would write your first step as:

(1/3)¹ =3^-1

As I understand it, the rule is 1) take the reciprocal of the base and 2) multiply the exponent by negative one.

I like your expansion of the second line.

Unfortunately, this question was just part of the confusion of a larger question. If folks are still willing to play...

The larger question involves two geometric sequences (164, 54, 18, ... and 2/81, 2/27/ 2/9, ...) that have their n^th term equal. What is the value of n?

I see that the first sequence decreases by a third each step while the second sequence increases by a third (again, each step). After that, I'm at a bit of a loss.

It's okay if I help the student for this problem. The correct answer is given. What I and the student don't thoroughly understand is all of the steps to get there.

thanks again, Todd's 5^th keyboard

Assuming that the first number of the first sequence is actually 162 rather than 164, the 5th terms of each sequence will be equal at a value of two.

162, 54, 18, 6, 2, ...
and
2/81, 2/27, 2/9, 2/3, 2/1 (or 2)...

In the first sequence the previous number is divided by three to get the next value. In the second sequence the previous number is multiplied by three to get the next value (or if you prefer, the previous denominator is divided by three).



Edited 1 time(s). Last edit at 09/28/2022 07:15PM by m.paris.

162/3ⁿ = (2/81)X3ⁿ
162=(2/81)X3²ⁿ
162X81/2=3²ⁿ
81X81=81²=3²ⁿ
3²ⁿ=81²
3ⁿ=81
n=4

Multiple edits because I skipped the steps that I consider intuitive. Note that 3ⁿX3ⁿ=3²ⁿ
and 162X81/2=(162/2)X81=81X81=81²
And the penultimate step is taking a square root (this gives you potential negative numbers as well, which should be irrelevant here).

This is how we learned to do it in my time. I know things may be stepped or annotated differently in US today, so I generally stayed away from helping my kids with their math.



Edited 4 time(s). Last edit at 09/29/2022 10:42AM by dk62.

And yes, your first equation is correct.

dk62,

Thanks for stepping up. Your answer is clear.

The answer key that comes with the course uses n-1 instead of just n.

The first line is:

162/3^n-1 = (2/81)(3^n-1)

I need to do some further digging to understand the differences between the two versions.

thanks again, Todd's still-learning board

That's because my solution counts the number of steps and not the number of items in the sequence, I.e., does not count the original starting number where the multiplier is 1, which is 3⁰. So by the time you reach equal numbers, by using 3 to the 4th, you have reached the 5th number in the sequence. My bad.



Edited 1 time(s). Last edit at 09/29/2022 06:37PM by dk62.

To make it simpler, you can just say n=x+1 and put x in all the equations above.