Geometric sequence and interest: Debt owed = ar^n - x{(r^n - 1)/(r - 1)}

A

Aminta_1900

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On the subject of geometric sequences, I am having difficulty with the second of these two questions. The first I have answered correctly using the formula given at the beginning of the section, namely (where x is the yearly installments to be paid back, a the principle loan, r the interest rate, and n the period)

Debt owed = ar^n - x{(r^n - 1)/(r - 1)}

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11) So 10,000(1.1)^n - 2,000{(1.1^n - 1)/(1.1 - 1)}

Using iteration to find n = 8 years (when the answer is < 0).

12) But in question 12 I am taking it that I have to now solve for x where n is given, but I am confused as to what the question means by "interest, calculated monthly, at 11% per annum", and so cannot discern what i should use for the period n.

I have been using n = 24 (for months) but this clearly wrong.

The answer is £23.31.

Thanks.
 
Aminta_1900 said:
On the subject of geometric sequences, I am having difficulty with the second of these two questions. The first I have answered correctly using the formula given at the beginning of the section, namely (where x is the yearly installments to be paid back, a the principle loan, r the interest rate, and n the period)

Debt owed = ar^n - x{(r^n - 1)/(r - 1)}

View attachment 35951

11) So 10,000(1.1)^n - 2,000{(1.1^n - 1)/(1.1 - 1)}

Using iteration to find n = 8 years (when the answer is < 0).

12) But in question 12 I am taking it that I have to now solve for x where n is given, but I am confused as to what the question means by "interest, calculated monthly, at 11% per annum", and so cannot discern what i should use for the period n.

I have been using n = 24 (for months) but this clearly wrong.

The answer is £23.31.

Thanks.
Click to expand...
According to your class-notes/textbook

What are the definitions of the variables 'r' and 'n' used in the given equation?​
 
Aminta_1900 said:
On the subject of geometric sequences, I am having difficulty with the second of these two questions. The first I have answered correctly using the formula given at the beginning of the section, namely (where x is the yearly installments to be paid back, a the principle loan, r the interest rate, and n the period)

Debt owed = ar^n - x{(r^n - 1)/(r - 1)}
Click to expand...
In this formula, r is not the interest rate; it would be 1 + the interest rate (per period). And n is not the period; it would be the number of periods. Am I right? This is how you used them in the first problem (#11).

So you are right that n = 24; but you have to calculate r based on the monthly rate, not the annual rate. That is your most likely error.

(On the side, your use of iteration to solve the first problem suggests that you either have not learned about logarithms, or didn't think about using them. That's what I would do for that one; it isn't needed for the second.)
 
Dr.Peterson said:
In this formula, r is not the interest rate; it would be 1 + the interest rate (per period). And n is not the period; it would be the number of periods. Am I right? This is how you used them in the first problem (#11).

So you are right that n = 24; but you have to calculate r based on the monthly rate, not the annual rate. That is your most likely error.

(On the side, your use of iteration to solve the first problem suggests that you either have not learned about logarithms, or didn't think about using them. That's what I would do for that one; it isn't needed for the second.)
Click to expand...

Dr.Peterson said:
In this formula, r is not the interest rate; it would be 1 + the interest rate (per period). And n is not the period; it would be the number of periods. Am I right? This is how you used them in the first problem (#11).

So you are right that n = 24; but you have to calculate r based on the monthly rate, not the annual rate. That is your most likely error.

(On the side, your use of iteration to solve the first problem suggests that you either have not learned about logarithms, or didn't think about using them. That's what I would do for that one; it isn't needed for the second.)
Click to expand...
OK thank you, I've got it now. For some reason I was seeing 11% as the product and not the sum of 12 months. I'm sure I even tried r = 1.009 at one point, but maybe not with n = 24.

I have learned about logarithms but nothing that would spring to mind when viewing question 11. I would only assume n * log {10,000(1.1)} = log [2,000{(1.1^n - 1)/(1.1 - 1)}] but wouldn't know where to go from there solving for n.

(I end up with 0.002n - 1 = log (1.1^n - 1) if I have done it correctly....)
 
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Hmmm, I get £23.18 as the monthly repayment. ?

But that was just a quick 'run through' and I've probably made a mistake somewhere. ?‍♂️?
 
The Highlander said:
If the annual interest rate is 11% then the corresponding monthly interest rate is:-

[math] (\sqrt[12]{1.11} -1)\times100\approx 0.8735\% \text{ per month}[/math]If that's of any help?
Click to expand...
Please remember "annual percentage rate (APR)" is different from "annual percentage yield (APY)".
 
Aminta_1900 said:
OK thank you, I've got it now. For some reason I was seeing 11% as the product and not the sum of 12 months. I'm sure I even tried r = 1.009 at one point, but maybe not with n = 24.

I have learned about logarithms but nothing that would spring to mind when viewing question 11. I would only assume n * log {10,000(1.1)} = log [2,000{(1.1^n - 1)/(1.1 - 1)}] but wouldn't know where to go from there solving for n.

(I end up with 0.002n - 1 = log (1.1^n - 1) if I have done it correctly....)
Click to expand...
Actually I see it is 0.04n = log {(1.1^n - 1)/0.5} ....but still may have rushed this.
 
monthly interest is (11%)/12 [imath]\implies r = \dfrac{0.11}{12}[/imath]

A = monthly payment, P = loan principle, n is 24

[imath]A = \dfrac{P \cdot r(1+r)^n}{(1+r)^n - 1}[/imath]
 
Aminta_1900 said:
I have learned about logarithms but nothing that would spring to mind when viewing question 11. I would only assume n * log {10,000(1.1)} = log [2,000{(1.1^n - 1)/(1.1 - 1)}] but wouldn't know where to go from there solving for n.

(I end up with 0.002n - 1 = log (1.1^n - 1) if I have done it correctly....)
Click to expand...
Aminta_1900 said:
Actually I see it is 0.04n = log {(1.1^n - 1)/0.5} ....but still may have rushed this.
Click to expand...
I assume this is all about question 11, for which you started with
Aminta_1900 said:
11) So 10,000(1.1)^n - 2,000{(1.1^n - 1)/(1.1 - 1)}

Using iteration to find n = 8 years (when the answer is < 0).
Click to expand...
But I can't tell what your steps were to get to these equations, neither of which is helpful.

Let's take it more slowly.

You want 10,000(1.1)^n = 2,000{(1.1^n - 1)/(1.1 - 1)}, right? That simplifies to

10,000(1.1)^n = 2,000{(1.1^n - 1)/(0.1)}

10,000(1.1)^n = 20,000{(1.1^n - 1)}

(1.1)^n = 2{(1.1^n - 1)}

(1.1)^n = 2(1.1^n) - 2

Do you see that the same exponential appears twice? Isolate that, and you can use a log.
 
Aminta_1900 said:
On the subject of geometric sequences, I am having difficulty with the second of these two questions. The first I have answered correctly using the formula given at the beginning of the section, namely (where x is the yearly installments to be paid back, a the principle loan, r the interest rate, and n the period)

Debt owed = ar^n - x{(r^n - 1)/(r - 1)}

View attachment 35951

11) So 10,000(1.1)^n - 2,000{(1.1^n - 1)/(1.1 - 1)}

Using iteration to find n = 8 years (when the answer is < 0).

12) But in question 12 I am taking it that I have to now solve for x where n is given, but I am confused as to what the question means by "interest, calculated monthly, at 11% per annum", and so cannot discern what i should use for the period n.

I have been using n = 24 (for months) but this clearly wrong.

The answer is £23.31.

Thanks.
Click to expand...
Did you solve these problems with correct answers?

If you did, please share your work in detail for completeness.
 
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