Geometric Progression - Pearson Core Mathematics 2

[Lorewill]

I have a problem in geometric progression chapter, Core Mathematics 2 A-Level, for Exedcel exam, of and it goes like this: the amount of 80000 should be reduced each year by 5000 but a rate of 4% is applied on the remaining debt, I am asked about the total time which the amount will be paid? (25y/book answer)

I believe if it were a linear progression it would go like this d=80000-500t, where d stands for debt and t, time in years. It is pretty clear that it would take 16 years if nothing is applied to initial amount. Like the physics equation vf=vi-at...

Following the rule I get this figures 78000 ((80000-5000)x1.04) for first year, 75920 ((78000-5000)x1.04) for the second year, the third year being 73756.80, fourth 71507.07 and so on. Dividing the second by the first, second by the third to find common ration I get something around 0.97... Would be right to set up something like d=80000(0.97...)^t or is there factor that could be applied to d=80000-5000t where a variable correction factor is added to 80000 so it increases in small steps? I believe I can not do something like log(0)=Log(80000(0.97...)^t) to find the time in years, there is no such thing as log(0)...

I research a equation to get that result in some business math tutorials, despite the lingo, they always have the time variable with them and that is precisely what I need to get to. This book has some dodge answer in the back but the majority is correct. I am also uploading a picture of it so you can actually read the problem as well.

 

[BigBeachBanana]

Before I provide any further explaination, I would like to know how familiar you are with the concept of present value and future value?

 

[blamocur]

My approach is to write out the first several terms of the sequence and try finding a pattern. Here is how I would approach your problem: let [imath]p_k[/imath] be the amount of the debt after the end of the [imath]k[/imath]-th year, [imath]a=5000[/imath] is the annual payment, [imath]\beta=1.04[/imath] is the "interest multiplier". We can write out the recursive formula: [imath]p_k = \beta (p_{k-1} - a)[/imath]. Then for the first several years we get this:
[math]p_0 = 80000[/math][math]p_1 = \beta p_0 - \beta a[/math][math]p_2 = \beta^2 p_0 - a(\beta^2 + \beta)[/math][math]p_3 = \beta^3 p_0 - a(\beta^3 + \beta^2 + \beta)[/math]Can you find a pattern there, then work out and prove a general formula for [imath]p_k[/imath] in terms of [imath]p_0, a, \beta[/imath] ?

 

[BigBeachBanana]

In financial mathematics theory, at time 0, your loan amount should be equal to the present value of your loan payments i.e:
[math]80,000= \underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}\\[/math]Can you continue?

 

[JeffM]

I have a problem in geometric progression chapter, Core Mathematics 2 A-Level, for Exedcel exam, of and it goes like this: the amount of 80000 should be reduced each year by 5000 but a rate of 4% is applied on the remaining debt, I am asked about the total time which the amount will be paid? (25y/book answer)

I believe if it were a linear progression it would go like this d=80000-500t, where d stands for debt and t, time in years. It is pretty clear that it would take 16 years if nothing is applied to initial amount. Like the physics equation vf=vi-at...

Following the rule I get this figures 78000 ((80000-5000)x1.04) for first year, 75920 ((78000-5000)x1.04) for the second year, the third year being 73756.80, fourth 71507.07 and so on. Dividing the second by the first, second by the third to find common ration I get something around 0.97... Would be right to set up something like d=80000(0.97...)^t or is there factor that could be applied to d=80000-5000t where a variable correction factor is added to 80000 so it increases in small steps? I believe I can not do something like log(0)=Log(80000(0.97...)^t) to find the time in years, there is no such thing as log(0)...

I research a equation to get that result in some business math tutorials, despite the lingo, they always have the time variable with them and that is precisely what I need to get to. This book has some dodge answer in the back but the majority is correct. I am also uploading a picture of it so you can actually read the problem as well.
View attachment 30436

First, that does not describe how a self-amortizing mortgage actually works. What normally is meant is this. Interest is computed annually at 4% and added to the amount due. Then an annual payment is deducted to compute the new balance due. Algebraically,

[math]D_{k+1} = D_k + 0.04 * D_k - C = 1.04D_k - C, \text { where}\\ D_k = \text {the amount due at the START of the } k^{th} \text { year, and}\\ C = \text {a constant payment made at the end of each year.}[/math]
With me so far?

The debt is to be paid off after n years.

So we have an equation

[math]D_{n+1} = 0 \implies 1.04D_n - C = 0 \implies[/math]
[math]1.04(1.04D_{n-1} - C) - C = 1.04^2D_{n-1} - C * \sum_{k=1}^2 1.04^{(k-1)} = 0 \implies[/math]
[math]D_1 * 1.04^n - C * \sum_{k=1}^n 1.04^{(k-1)} = 0 \implies[/math]
[math]D_1 * 1.04^n = C * \dfrac{1.04^n - 1}{0.04} \implies[/math]
[math]\dfrac{0.04D_1* 1.04^n}{C} = 1.04^n - 1 \implies \dfrac{0.04D_1}{C} = 1 - \dfrac{1}{1.04^n} \implies [/math]
[math]\left ( \dfrac{1}{1.04}\right )^n = \dfrac{C - 0.04D_1}{C} \implies log \left\{ \left ( \dfrac{1}{1.04}\right )^n \right \} = log \left (\dfrac{C - 0.04D_1}{C} \right) \implies[/math]
[math]n * log \left ( \dfrac{1}{1.04} \right) = log \left (\dfrac{C - 0.04D_1}{C} \right) \implies[/math]
[math]n = log \left (\dfrac{C - 0.04D_1}{C} \right) \div log \left (\dfrac{1}{1.04} \right) .[/math]
With me to here? But D_1 is 80000 and C is 5000. So

[math]n = \dfrac{log \left ( \dfrac{5000 - 0.04 * 80000}{5000} \right )} {log \left ( \dfrac{1}{1.04} \right )} \approx 26.[/math]
How did your book come up with 25? They did not charge any interest for the first year! Totally weird. Few if any banks will lend you 80000 pounds for a year at 0 interest. But you can work out the formula that the book used to get its answer if you use my work as a model.

Oh, and at least in the US, this is all done monthly rather than annually.

 

[Lorewill]

My approach is to write out the first several terms of the sequence and try finding a pattern. Here is how I would approach your problem: let [imath]p_k[/imath] be the amount of the debt after the end of the [imath]k[/imath]-th year, [imath]a=5000[/imath] is the annual payment, [imath]\beta=1.04[/imath] is the "interest multiplier". We can write out the recursive formula: [imath]p_k = \beta (p_{k-1} - a)[/imath]. Then for the first several years we get this:
[math]p_0 = 80000[/math][math]p_1 = \beta p_0 - \beta a[/math][math]p_2 = \beta^2 p_0 - a(\beta^2 + \beta)[/math][math]p_3 = \beta^3 p_0 - a(\beta^3 + \beta^2 + \beta)[/math]Can you find a pattern there, then work out and prove a general formula for [imath]p_k[/imath] in terms of [imath]p_0, a, \beta[/imath] ?

I believe I had try on that style. The answer I had gotten was like 70 years. My approach is to write out the first several terms of the sequence and try finding a pattern. Here is how I would approach your problem: let [imath]p_k[/imath] be the amount of the debt after the end of the [imath]k[/imath]-th year, [imath]a=5000[/imath] is the annual payment, [imath]\beta=1.04[/imath] is the "interest multiplier". We can write out the recursive formula: [imath]p_k = \beta (p_{k-1} - a)[/imath]. Then for the first several years we get this:

[math]p_0 = 80000[/math][math]p_1 = \beta p_0 - \beta a[/math][math]p_2 = \beta^2 p_0 - a(\beta^2 + \beta)[/math][math]p_3 = \beta^3 p_0 - a(\beta^3 + \beta^2 + \beta)[/math]Can you find a pattern there, then work out and prove a general formula for [imath]p_k[/imath] in terms of [imath]p_0, a, \beta[/imath] ?

First, that does not describe how a self-amortizing mortgage actually works. What normally is meant is this. Interest is computed annually at 4% and added to the amount due. Then an annual payment is deducted to compute the new balance due. Algebraically,

[math]D_{k+1} = D_k + 0.04 * D_k - C = 1.04D_k - C, \text { where}\\ D_k = \text {the amount due at the START of the } k^{th} \text { year, and}\\ C = \text {a constant payment made at the end of each year.}[/math]
With me so far?

The debt is to be paid off after n years.

So we have an equation

[math]D_{n+1} = 0 \implies 1.04D_n - C = 0 \implies[/math]
[math]1.04(1.04D_{n-1} - C) - C = 1.04^2D_{n-1} - C * \sum_{k=1}^2 1.04^{(k-1)} = 0 \implies[/math]
[math]D_1 * 1.04^n - C * \sum_{k=1}^n 1.04^{(k-1)} = 0 \implies[/math]
[math]D_1 * 1.04^n = C * \dfrac{1.04^n - 1}{0.04} \implies[/math]
[math]\dfrac{0.04D_1* 1.04^n}{C} = 1.04^n - 1 \implies \dfrac{0.04D_1}{C} = 1 - \dfrac{1}{1.04^n} \implies [/math]
[math]\left ( \dfrac{1}{1.04}\right )^n = \dfrac{C - 0.04D_1}{C} \implies log \left\{ \left ( \dfrac{1}{1.04}\right )^n \right \} = log \left (\dfrac{C - 0.04D_1}{C} \right) \implies[/math]
[math]n * log \left ( \dfrac{1}{1.04} \right) = log \left (\dfrac{C - 0.04D_1}{C} \right) \implies[/math]
[math]n = log \left (\dfrac{C - 0.04D_1}{C} \right) \div log \left (\dfrac{1}{1.04} \right) .[/math]
With me to here? But D_1 is 80000 and C is 5000. So

[math]n = \dfrac{log \left ( \dfrac{5000 - 0.04 * 80000}{5000} \right )} {log \left ( \dfrac{1}{1.04} \right )} \approx 26.[/math]
How did your book come up with 25? They did not charge any interest for the first year! Totally weird. Few if any banks will lend you 80000 pounds for a year at 0 interest. But you can work out the formula that the book used to get its answer if you use my work as a model.

Oh, and at least in the US, this is all done monthly rather than annually.

Thank you for the response. I believe it is all about coming with a expression more like vf=vo+at style. Most of demonstration so far have been simple. I believe the question was there for the use of geometric progression concepts. I have the idea of how you set the answer but I don't have the experience or the training to read it well, to be honest. There were some question on analytical geometry that I had to research but were later introduced in other chapters. It doesn't feel a put well together book but overall it seems designed as introduction to this maths. I have never bump into any book that would actually help do logic, deductive work...

 

[Lorewill]

My approach is to write out the first several terms of the sequence and try finding a pattern. Here is how I would approach your problem: let [imath]p_k[/imath] be the amount of the debt after the end of the [imath]k[/imath]-th year, [imath]a=5000[/imath] is the annual payment, [imath]\beta=1.04[/imath] is the "interest multiplier". We can write out the recursive formula: [imath]p_k = \beta (p_{k-1} - a)[/imath]. Then for the first several years we get this:
[math]p_0 = 80000[/math][math]p_1 = \beta p_0 - \beta a[/math][math]p_2 = \beta^2 p_0 - a(\beta^2 + \beta)[/math][math]p_3 = \beta^3 p_0 - a(\beta^3 + \beta^2 + \beta)[/math]Can you find a pattern there, then work out and prove a general formula for [imath]p_k[/imath] in terms of [imath]p_0, a, \beta[/imath] ?

From the answer I get I understand yours better. I had tried the same approach but when I didn't manage to set an convergent sums toward zero, if I remember. I don't have the skill set to make it equal to zero I guess.

 

[Lorewill]

Before I provide any further explaination, I would like to know how familiar you are with the concept of present value and future value?

I researched it on-line but the exercise is lodge in a chapter about geometric progression.

 

[Lorewill]

In financial mathematics theory, at time 0, your loan amount should be equal to the present value of your loan payments i.e:
[math]80,000= \underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}\\[/math]Can you continue?

I understand it but I don't have much experience with that type of calculation. I think I should (1.04)^-n and build geometric sum out of it, am I write? I am giving it a try them but I am more familiar with linear and quadratic equation. I haven't had much practice on reasoning but I think a lot people if not the majority of people are on the same boat as well.

 

[BigBeachBanana]

I understand it but I don't have much experience with that type of calculation. I think I should (1.04)^-n and build geometric sum out of it, am I write? I am giving it a try them but I am more familiar with linear and quadratic equation. I haven't had much practice on reasoning but I think a lot people if not the majority of people are on the same boat as well.

Here's the sum of a finite geometric series:


What's "a" and "r" of the sum below?
[math]\underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}[/math]

 

[JeffM]

I believe I had try on that style. The answer I had gotten was like 70 years. My approach is to write out the first several terms of the sequence and try finding a pattern. Here is how I would approach your problem: let [imath]p_k[/imath] be the amount of the debt after the end of the [imath]k[/imath]-th year, [imath]a=5000[/imath] is the annual payment, [imath]\beta=1.04[/imath] is the "interest multiplier". We can write out the recursive formula: [imath]p_k = \beta (p_{k-1} - a)[/imath]. Then for the first several years we get this:



Thank you for the response. I believe it is all about coming with a expression more like vf=vo+at style. Most of demonstration so far have been simple. I believe the question was there for the use of geometric progression concepts. I have the idea of how you set the answer but I don't have the experience or the training to read it well, to be honest. There were some question on analytical geometry that I had to research but were later introduced in other chapters. It doesn't feel a put well together book but overall it seems designed as introduction to this maths. I have never bump into any book that would actually help do logic, deductive work...

I am sorry. Financial formulas involving periodic payments are not as simple as that. My very first job working in a bank was making sure that the computers were calculating interest and payments correctly. It is not rocket science, but it is a bit more involved than the formula for instantaneous velocity.

Please tell me where you first run into difficulty with my logic and why, and I can explain more fully.

The basic equation is [imath]D_1(1 + r)^n - C * \dfrac{1 - (1 + r)^n}{1 - (1 + r)} = 0[/imath], which you should recognize as containing a geometric series. (As I explained in my first post, that formula will not work for this problem exactly because they are assuming that the bank does not charge interest for the first year).

If you want to solve for n, you need logarithms as you initially suggested.

That basic equation simplifies, using basic algebra, to

[math]D_1(1 + r)^n = C * \dfrac{1 - (1 + r)^n}{1 - (1 + r)} \implies[/math]
[math]D_1(1 + r)^n = C * \dfrac{1 - (1 + r)^n}{-r} \implies[/math]
[math]-rD_1(1 + r)^n = C\{(1 - (1 + r)^n\} \implies[/math]
[math]-rD_1 = C \left (\dfrac{1}{(1 + r)^n} - 1 \right ) = C * \dfrac{1}{(1 + r)^n} - C \implies[/math]
[math]C - rD_1 = C * \left ( \dfrac{1}{1 + r)^n} \right ) = C * \left ( \dfrac{1}{1 + r} \right )^n \implies[/math]
[math]\dfrac{C - rD_1}{C} = \left ( \dfrac{1}{1 + r)} \right )^n.[/math]
To solve for n, we use logarithms.

[math]\dfrac{C - rD_1}{C} = \left ( \dfrac{1}{1 + r)} \right )^n \implies log \left (\dfrac{C - rD_1}{C} \right ) = log \left \{ \left ( \dfrac{1}{1 + r)} \right )^n\right \} \implies[/math]
[math]log \left ( \dfrac{C - rD_1}{C} \right ) = n * log \left ( \dfrac{1}{1 + r)} \right ) \implies[/math]
[math]n = \dfrac{log \left ( \dfrac{C - rD_1}{C} \right )}{log \left ( \dfrac{1}{1 + r)} \right )}.[/math]
Now tell me where you are getting lost. I am sorry that it is not a nice simple formula, but the universe does not always lead to simplicity.

 

[BigBeachBanana]

There are two ways to calculate the loan's outstanding balance in the mortgage business, namely the prospective method and the retrospective method. Both methods are derived using the sum geometric series as the nature of present value, and future values are the sum of geometric series. What @JeffM and @blamocur suggested is the retrospective method, and I suggest the prospective method. Here's how those are derived and the intuition behind them.

Retrospective method:
Ultimately, if you followed @blamocur's suggestion, this is the result.
[math]B_n=L(1+i)^n-P(\frac{(1+i)^n-1}{i})[/math]Where [imath]B_n[/imath] is the outstanding balance after n years, L is the principal amount, P is the repayment amount, and i is the annual interest rate.
In your problem, your goal is to solve for the when the outstanding balance =0 after n years, so
[math]0=80,000(1+.04)^n-5,000(\frac{(1+0.04)^n-1}{.04})\Rightarrow n \approx 26[/math]The intuition is as follows: The outstanding balance at n years is the same as your original loan amount of 80,000 accumulated interest for n years (as if you haven't paid back anything) subtract the accumulated value of all of the repayments of 5,000 to time n (adjusting for interest).

Prospective method:
[math]B_n=P(\frac{1-(1+i)^{-n}}{i})[/math]The intuition is as follows:

At time 0, your loan amount should be equal to the present value of your loan payments.
[math]B_0=80,000= \underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}=5,000(1.04)^{-1}(\frac{1-(1.04)^{-n}}{1-(1.04)^{-1}}) \Rightarrow n\approx 26[/math]
Here's a diagram of both methods:

Either method you use, you should get the same answer. Sometimes one is better than the other, depending on what is given.

 

[Lorewill]

Here's the sum of a finite geometric series:

View attachment 30451
What's "a" and "r" of the sum below?
[math]\underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}[/math]

The Greek letter represents a sum. The letter k indicates the initial input. The letter a the first term. I know the derivation of that formula and the Grauss too. My first approach was to do like Blocomur but somehow I did not manage to get it converging. I can read that better than the other solutions. As I was failing at that one, my idea was to use the equation from physics that vf=vi-at but adding an correcting factor to it and make the the "vf" equal to zero. I think that is more in line with type of question are within the book and for current level it is supposed to address, introductory, go and grab... I might have mounted it wrong because I got 70 years.

 

[Lorewill]

There are two ways to calculate the loan's outstanding balance in the mortgage business, namely the prospective method and the retrospective method. Both methods are derived using the sum geometric series as the nature of present value, and future values are the sum of geometric series. What @JeffM and @blamocur suggested is the retrospective method, and I suggest the prospective method. Here's how those are derived and the intuition behind them.

Retrospective method:
Ultimately, if you followed @blamocur's suggestion, this is the result.
[math]B_n=L(1+i)^n-P(\frac{(1+i)^n-1}{i})[/math]Where [imath]B_n[/imath] is the outstanding balance after n years, L is the principal amount, P is the repayment amount, and i is the annual interest rate.
In your problem, your goal is to solve for the when the outstanding balance =0 after n years, so
[math]0=80,000(1+.04)^n-5,000(\frac{(1+0.04)^n-1}{.04})\Rightarrow n \approx 26[/math]The intuition is as follows: The outstanding balance at n years is the same as your original loan amount of 80,000 accumulated interest for n years (as if you haven't paid back anything) subtract the accumulated value of all of the repayments of 5,000 to time n (adjusting for interest).

Prospective method:
[math]B_n=P(\frac{1-(1+i)^{-n}}{i})[/math]The intuition is as follows:

At time 0, your loan amount should be equal to the present value of your loan payments.
[math]B_0=80,000= \underbrace{5,000(1.04)^{-1}+5,000(1.04)^{-2}+...+5,000(1.04)^{-n}}_{\text{finite geometric series}}=5,000(\frac{1-(1.04)^{-(n+1)}}{1-(1.04)^{-1}}) \Rightarrow n\approx 26[/math]
Here's a diagram of both methods:
View attachment 30452
Either method you use, you should get the same answer. Sometimes one is better than the other, depending on what is given.

Thank you for answer. I will return to Blacomur method which I had worked on somehow and see what I have done wrong with it... If I still bumping into walls, I return for help. Thanks for your time.

 

[BigBeachBanana]

Thank you for answer. I will return to Blacomur method which I had worked on somehow and see what I have done wrong with it... If I still bumping into walls, I return for help. Thanks for your time.

If you show us your work, we can help you see what went wrong.

 

[Lorewill]

I am sorry. Financial formulas involving periodic payments are not as simple as that. My very first job working in a bank was making sure that the computers were calculating interest and payments correctly. It is not rocket science, but it is a bit more involved than the formula for instantaneous velocity.

Please tell me where you first run into difficulty with my logic and why, and I can explain more fully.

The basic equation is [imath]D_1(1 + r)^n - C * \dfrac{1 - (1 + r)^n}{1 - (1 + r)} = 0[/imath], which you should recognize as containing a geometric series. (As I explained in my first post, that formula will not work for this problem exactly because they are assuming that the bank does not charge interest for the first year).

If you want to solve for n, you need logarithms as you initially suggested.

That basic equation simplifies, using basic algebra, to

[math]D_1(1 + r)^n = C * \dfrac{1 - (1 + r)^n}{1 - (1 + r)} \implies[/math]
[math]D_1(1 + r)^n = C * \dfrac{1 - (1 + r)^n}{-r} \implies[/math]
[math]-rD_1(1 + r)^n = C\{(1 - (1 + r)^n\} \implies[/math]
[math]-rD_1 = C \left (\dfrac{1}{(1 + r)^n} - 1 \right ) = C * \dfrac{1}{(1 + r)^n} - C \implies[/math]
[math]C - rD_1 = C * \left ( \dfrac{1}{1 + r)^n} \right ) = C * \left ( \dfrac{1}{1 + r} \right )^n \implies[/math]
[math]\dfrac{C - rD_1}{C} = \left ( \dfrac{1}{1 + r)} \right )^n.[/math]
To solve for n, we use logarithms.

[math]\dfrac{C - rD_1}{C} = \left ( \dfrac{1}{1 + r)} \right )^n \implies log \left (\dfrac{C - rD_1}{C} \right ) = log \left \{ \left ( \dfrac{1}{1 + r)} \right )^n\right \} \implies[/math]
[math]log \left ( \dfrac{C - rD_1}{C} \right ) = n * log \left ( \dfrac{1}{1 + r)} \right ) \implies[/math]
[math]n = \dfrac{log \left ( \dfrac{C - rD_1}{C} \right )}{log \left ( \dfrac{1}{1 + r)} \right )}.[/math]
Now tell me where you are getting lost. I am sorry that it is not a nice simple formula, but the universe does not always lead to simplicity.

I think I understand the first lay out of the first post you published. I think I was bit lost when sum sign popped up and move to the general formula to resolve the question. My lack of expertise and experience I guess. I believe the very first instalment was in with the correcting factor on something vf=vi-at on vi which my lack skills prevent from getting it right.

 

[JasonTurner]

The math problem here presents a problematic situation, which is to be expressed in different linear expressions and then solved. The statement is about some financial issues related to the mortgage and interest. The total time the entire mortgage is paid is interest in mind. The appeal of the 2nd year is 4%, and the total mortgage is calculated to be £80,000, and payment is to be made annually at the rate of £5000.
One of the possible solutions starts with listing several equations, which will help eliminate possible equations length and ease the answer. Numbering the equations will help ease out more, and using the equations in each other will help solve the time variable through these equations.
In the second case of a solution, the solver creates a set of equations that uses different variables to calculate the period for which mortgage to be paid. Through a series of steps and using values for stated variables, the period for the mortgage comes to be 26 years a year extra than the book answer.