Yearly loan repayments from C = A/1+i + A/(1+i)^2 + A/(1+i)^3 + A/(1+i)^4 + A/(1+i)^5

[Simonsky]

Given this formula for a five year repayment plan where C= Amount borrowed; A = amount repaid each year; i = annual interest rate (AIR) expressed as a decimal:

C = A/1+i + A/(1+i)^2 + A/(1+i)^3 + A/(1+i)^4 + A/(1+i)^5

If they borrow £30,000 how much will each yearly repayment be at an AIR of 6%?

So 30 000 = X/1.06 + x/1.06^2 + x/1.06^3 + x/1.06^4 + x/1.06^5

but I'm not sure how to proceed here- there's probably a simple method I can't get at-seem to be having 'brain freeze' on this.

Any pointers appreciated.

 

[Dr.Peterson]

Given this formula for a five year repayment plan where C= Amount borrowed; A = amount repaid each year; i = annual interest rate (AIR) expressed as a decimal:

C = A/1+i + A/(1+i)^2 + A/(1+i)^3 + A/(1+i)^4 + A/(1+i)^5

If they borrow £30,000 how much will each yearly repayment be at an AIR of 6%?

So 30 000 = X/1.06 + x/1.06^2 + x/1.06^3 + x/1.06^4 + x/1.06^5

but I'm not sure how to proceed here- there's probably a simple method I can't get at-seem to be having 'brain freeze' on this.

Any pointers appreciated.

Factor out the x on the right-hand side!

There are other ways to do this, such as applying the formula for a geometric series if you know about that, but that is not absolutely essential.

 

[Simonsky]

Factor out the x on the right-hand side!

There are other ways to do this, such as applying the formula for a geometric series if you know about that, but that is not absolutely essential.

Thanks for reply Dr. Peterson.

factor out the X -hmmm..... not sure- 1/X( 1.06 etc) that doesn't seem right...

Still got brain freeze on this, sorry..

 

[Dr.Peterson]

Thanks for reply Dr. Peterson.

factor out the X -hmmm..... not sure- 1/X( 1.06 etc) that doesn't seem right...

Still got brain freeze on this, sorry..

Don't forget that, say, x/1.06 means the same thing as x * 1/1.06. Rewrite each term that way before factoring.

 

[Simonsky]

Don't forget that, say, x/1.06 means the same thing as x * 1/1.06. Rewrite each term that way before factoring.

many thanks-I think I get it (hopefully):

So: we get x(1/1.06+1/1.06^2 + 1/1.06^3 + 1/1.06^4 + 1/1.06^5) = 4.21X (rounded) = 30,000

So x + 7126 ( to four sig. fig.) Answer in book says 7121.89 so I've lost a bit of accuracy there somehow-perhaps I rounded down too much.

 

[Dr.Peterson]

many thanks-I think I get it (hopefully):

So: we get x(1/1.06+1/1.06^2 + 1/1.06^3 + 1/1.06^4 + 1/1.06^5) = 4.21X (rounded) = 30,000

So x = 7126 ( to four sig. fig.) Answer in book says 7121.89 so I've lost a bit of accuracy there somehow-perhaps I rounded down too much.

That's correct. I get their answer when I enter the whole expression into the calculator at once, giving 4.21236... and then divide.

Generally speaking, you should never round intermediate calculations more than necessary; and if I knew I would need six significant digits in my answer (to be accurate to the penny), I would be sure to keep at least that much precision in my intermediate results if I had to round. But the main point is that you were able to solve it!

 

[Simonsky]

That's correct. I get their answer when I enter the whole expression into the calculator at once, giving 4.21236... and then divide.

Generally speaking, you should never round intermediate calculations more than necessary; and if I knew I would need six significant digits in my answer (to be accurate to the penny), I would be sure to keep at least that much precision in my intermediate results if I had to round. But the main point is that you were able to solve it!

Well..with your help-which is much appreciated.

The next question is a sort of reversal of the above one: If they borrow 30000 but cannot afford to repay more than 8000 a year what is the maximum interest rate they can borrow at?

So 30000 = 8000/1+X + 8000/1+x^2 + 8000/1+X^3 +8000/1+X^4 +8000/1+X^5

Hmm...not sure what to do next-this stuff is really flooring me at present.....

 

[Dr.Peterson]

If they borrow 30000 but cannot afford to repay more than 8000 a year what is the maximum interest rate they can borrow at?

So 30000 = 8000/1+X + 8000/1+x^2 + 8000/1+X^3 +8000/1+X^4 +8000/1+X^5

Hmm...not sure what to do next-this stuff is really flooring me at present.....

You mean this:

30000 = 8000/(1+x) + 8000/(1+x)^2 + 8000/(1+x)^3 +8000/(1+x)^4 +8000/(1+x)^5
​

The parentheses are essential! (And variables x and X are not considered the same in algebra.)

Anyway, I mentioned in passing previously the formula for the sum of a (finite) geometric series, though it wasn't necessary then. Now it is. So I have to ask: Do you know anything about that?

Even then, I'm not sure this can be solved algebraically. (I haven't done it yet.) If that's true, and this was assigned to you, we need to know not only what you know about series, but also whether you can use technology (such as a graphing calculator or financial calculator) to get numerical solutions, or have been taught any such methods. Anything about your context may help us know how to help.

 

[Simonsky]

You mean this:

30000 = 8000/(1+x) + 8000/(1+x)^2 + 8000/(1+x)^3 +8000/(1+x)^4 +8000/(1+x)^5
​

The parentheses are essential! (And variables x and X are not considered the same in algebra.)

Anyway, I mentioned in passing previously the formula for the sum of a (finite) geometric series, though it wasn't necessary then. Now it is. So I have to ask: Do you know anything about that?

Even then, I'm not sure this can be solved algebraically. (I haven't done it yet.) If that's true, and this was assigned to you, we need to know not only what you know about series, but also whether you can use technology (such as a graphing calculator or financial calculator) to get numerical solutions, or have been taught any such methods. Anything about your context may help us know how to help.

Probably best if I show you the page in the book - it's a book for what we call GCSE in the UK an exam taken at age 16 (I'm...ahem... somewhat older than that but just trying to become math(s) literate).

The finite geometric series formula as I've learnt it in the past is: Ak+1 =RAk (R is common ratio) but that isn't taught at this level.

Anyway best if I show you the page:



It's part of 'Task 5' using the five year loan formula above.

 

[Dr.Peterson]

Probably best if I show you the page in the book - it's a book for what we call GCSE in the UK an exam taken at age 16 (I'm...ahem... somewhat older than that but just trying to become math(s) literate).

The finite geometric series formula as I've learnt it in the past is: A_k+1 =RA_k (R is common ratio) but that isn't taught at this level.

It's part of 'Task 5' using the five year loan formula above.

The formula you show is actually the definition of a geometric sequence, not a formula for a series. (A sequence is a list of numbers; a series is a summation of terms in a sequence.) There is a formula for the sum of a geometric series, which part (d) implies you have not learned. (I almost showed you the formula in (d) last time, but chose to hold off. It is formed by applying the series formula I mentioned to this specific problem.)

Unfortunately, the page doesn't give me any idea how they expect you to answer (c). But it seems clear that they don't expect you to know any fancy methods! Possibly all they want you to do is to use the results from (b) to decide whether 6% or 8% could be the answer, and perhaps try 7% or other values in a guess-and-check approach.

If I were tutoring you face to face, I would be taking the book from you and looking through it to get a feel for the level of knowledge they expect of you, and whether previous examples might have prepared you to solve a problem that can't be done by algebra. (Many textbooks fail to tell students that algebra can't solve everything, so this one may well be filling that gap by including, without comment, problems of this sort. But I would hope that they would occasionally warn you to expect this!)

 

[Simonsky]

The formula you show is actually the definition of a geometric sequence, not a formula for a series. (A sequence is a list of numbers; a series is a summation of terms in a sequence.) There is a formula for the sum of a geometric series, which part (d) implies you have not learned. (I almost showed you the formula in (d) last time, but chose to hold off. It is formed by applying the series formula I mentioned to this specific problem.)

Unfortunately, the page doesn't give me any idea how they expect you to answer (c). But it seems clear that they don't expect you to know any fancy methods! Possibly all they want you to do is to use the results from (b) to decide whether 6% or 8% could be the answer, and perhaps try 7% or other values in a guess-and-check approach.

If I were tutoring you face to face, I would be taking the book from you and looking through it to get a feel for the level of knowledge they expect of you, and whether previous examples might have prepared you to solve a problem that can't be done by algebra. (Many textbooks fail to tell students that algebra can't solve everything, so this one may well be filling that gap by including, without comment, problems of this sort. But I would hope that they would occasionally warn you to expect this!)

I think you are right. The question probably expects you to use the answer ti b) and do 'guesstimates.' I think I tried to find an algebraic way because I have done more advanced maths in the past but forgotten most of it. Clearly, this book just expects you to try out values in the formula.

 

[Simonsky]

I think you are right. The question probably expects you to use the answer ti b) and do 'guesstimates.' I think I tried to find an algebraic way because I have done more advanced maths in the past but forgotten most of it. Clearly, this book just expects you to try out values in the formula.

Having said that, the answer given is 10.4% and that is not a value you are likely to arrive at by guessing - so maybe they are expecting a technique? it's weird because the book up to this point takes things by step and then here, it suddenly plunges you in the deep end without a lifebelt! The book is, of course, designed for school classroom use and I'm tutoring myself which is not ideal.

 

[Simonsky]

That is correct. 10.4248... which they rounded to 10.4

What method did you use to get there?

 

[JeffM]

What method did you use to get there?

There is a formula. Just giving you the formula will not advance you much. I'll derive it for you. May take a while.

 

[Dr.Peterson]

Having said that, the answer given is 10.4% and that is not a value you are likely to arrive at by guessing - so maybe they are expecting a technique? it's weird because the book up to this point takes things by step and then here, it suddenly plunges you in the deep end without a lifebelt! The book is, of course, designed for school classroom use and I'm tutoring myself which is not ideal.

That is strange. At the least they should have said "to the nearest tenth of a percent" so you don't have to keep improving the estimate forever.

The problem can in a sense be solved algebraically -- it becomes a 5th degree polynomial equation, and some such equations can be solved. But none of the methods I know other than graphing yield a solution. And standard methods of numerical solution would be beyond these students.

Are you sure the book hasn't had any examples where something like a graphing calculator is used? That's the only thing I can think of that is at the appropriate level.

 

[JeffM]

I am going to do this in pieces.

Do you know this theorem.

\(\displaystyle \displaystyle n \in \mathbb Z \text { and } n > 0 \implies a^{(n+1)} - b^{(n+1)} = (a - b) * \sum_{j=0}^na^{(n-j)}b^j.\)

If not, here is a proof.

\(\displaystyle n = 1 \implies a^{(1+1)} - b^{(1+1)} = a^2 - b^2 = (a - b)(a + b) = (a - b)(a^1b^0 + a^0b^1).\)

So suppose n > 1.

\(\displaystyle \displaystyle (a - b) * \sum_{j=0}^na^{(n-j)}b^j = a \left ( \sum_{j=0}^na^{(n-j)}b^j \right ) - b \left ( \sum_{j=0}^na^{(n-j)}b^j \right) =\)

\(\displaystyle \displaystyle \left ( \sum_{j=0}^na^{(n+1-j)}b^j \right ) - \left ( \sum_{j=0}^na^{(n-j)}b^{(j+1)} \right ) = \left ( \sum_{k=0}^na^{(n+1-k)}b^k \right ) - \left ( \sum_{k=1}^{n+1}a^{(n+1-k)}b^k \right ) = \)

\(\displaystyle \displaystyle \left ( a^{(n+1-0)}b^0 + \sum_{k=1}^na^{(n+1-k)}b^k \right ) - \left ( a^{(n+1-n-1)}b^{(n+1)} + \sum_{k=1}^na^{(n+1-k)}b^k \right ) = \)

\(\displaystyle \displaystyle a^{(n+1)} - b^{(n+1)} + \left ( \sum_{k=1}^na^{(n+1-k)}b^k \right ) - \left ( \sum_{k=1}^na^{(n+1-k)}b^k \right ) = a^{(n+1)} - b^{(n+1)}. \)

Now as a result of that theorem we get

\(\displaystyle \displaystyle a^{(n+1)} - 1 = (a - 1) * \sum_{j=0}^n a^j.\) And so

\(\displaystyle a \ne 1 \implies \displaystyle \sum_{j=0}^n a^j = \dfrac{a^{(n+1)} - 1}{a - 1}.\)

Tune in tomorrow for the next thrilling installment.

 

[Simonsky]

That is strange. At the least they should have said "to the nearest tenth of a percent" so you don't have to keep improving the estimate forever.

The problem can in a sense be solved algebraically -- it becomes a 5th degree polynomial equation, and some such equations can be solved. But none of the methods I know other than graphing yield a solution. And standard methods of numerical solution would be beyond these students.

Are you sure the book hasn't had any examples where something like a graphing calculator is used? That's the only thing I can think of that is at the appropriate level.

No-even the use of a graphing calculator is beyond the level of this text book-certainly dealing with 5th degree polynomials is well beyond this book which barely touches on calculus and only reaches simple quadratic equations. It's weird that it throws in more advanced stuff without any preparation-these questions are probably what we, in the UK would call 'A' level which is pre-university stuff. I might contact the publishers to see if they can let me contact the authors as it is a fairly recent publication fro schools.

it seems then, from what you say that there is no straight forward way of getting that 10.4 answer using algebra as we used to work out the yearly payments. Mystifying!