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Geometric Progressions

This is a dumb question, but idc. :smile:

In geometric investment type problems, what is the meaning of interest? also how do you calculate common ratio will be when given an interest percentage.

This question:
https://gyazo.com/ce01cfbe03619d817460aa4d760a63fd
You can think of it as a geometric progression of sorts, with A being the first term and the added interest being the "common ratio" (be aware that this is greater than 1 since interest adds to the initial investment) :smile:
(edited 8 years ago)
Reply 2
Original post by Naruke
This is a dumb question, but idc. :smile:

In geometric investment type problems, what is the meaning of interest? also how do you calculate common ratio will be when given an interest percentage.

This question:
https://gyazo.com/ce01cfbe03619d817460aa4d760a63fd


Interest of 4%, for instance, means that after a year, in this case, could be different unit of time, that his cash increases by 4%, or if you prefer by a factor of 1.04 (what he already has is 100% then you add 4% which is 4/100 or 0.04)
Reply 3
Original post by 1 8 13 20 42
Interest of 4%, for instance, means that after a year, in this case, could be different unit of time, that his cash increases by 4%, or if you prefer by a factor of 1.04 (what he already has is 100% then you add 4% which is 4/100 or 0.04)


Hi, again! Could you tell me where I've gone wrong?

https://gyazo.com/0565b8741b3035479502fc8a518b2acd

multiplication factor would be 100 % - 15% = 85%

85 % = 0.85

a(0.85)3=11054.25 a (0.85)^3 = 11054.25

a=11054.25(0.85)3 a = \frac{11054.25}{(0.85)^3}

Unparseable latex formula:

18000 (0.85)^n^-^1 < 5000



Unparseable latex formula:

(0.85)^n^-^1 < \frac {5}{18}



Unparseable latex formula:

log (0.85)^n^-^1 < log \frac {5}{18}



Unparseable latex formula:

n-1 < \frac {log \{5}{18}}{log 0.85)



n>8.881... n > 8.881 ...
Reply 4
Original post by Naruke
Hi, again! Could you tell me where I've gone wrong?

https://gyazo.com/0565b8741b3035479502fc8a518b2acd

multiplication factor would be 100 % - 15% = 85%

85 % = 0.85

a(0.85)3=11054.25 a (0.85)^3 = 11054.25

a=11054.25(0.85)3 a = \frac{11054.25}{(0.85)^3}

Unparseable latex formula:

18000 (0.85)^n^-^1 < 5000



Unparseable latex formula:

(0.85)^n^-^1 < \frac {5}{18}



Unparseable latex formula:

log (0.85)^n^-^1 < log \frac {5}{18}



Unparseable latex formula:

n-1 < \frac {log \{5}{18}}{log 0.85)



n>8.881... n > 8.881 ...


What is the answer? Didn't sleep last night so a bit slow but nothing jumps out at first
Reply 5
Original post by 1 8 13 20 42
What is the answer? Didn't sleep last night so a bit slow but nothing jumps out at first


You and me both, I woke up yesterday at 3pm and haven't slept since :biggrin:

I think the part I'm getting wrong is using

Unparseable latex formula:

(0.85)^n^-^1 < \frac {5}{18}



Unparseable latex formula:

log (0.85)^n^-^1 < log \frac {5}{18}



solutionbank just uses n, do you know why this is?
(edited 8 years ago)
Reply 6
Original post by Naruke
You and me both, I woke up yesterday at 3pm and haven't slept since :biggrin:

I think the part I'm getting wrong is using

Unparseable latex formula:

(0.85)^n^-^1 < \frac {5}{18}



Unparseable latex formula:

log (0.85)^n^-^1 < log \frac {5}{18}



solutionbank just uses n, do you know why this is?


I did think that might be erroneous, but then I thought I was missing something in thinking there was a problem..
but on closer inspection basically there's no need to use n-1; you see that after 7.88ish years the car will be at 5000, it seems you're going from year 1 instead of year 0 or something

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