1. ## Geometric Series Help

Hello,
I have a math problem that I cannot do, I have spent the past 3 hours on it and got relatively no where!!

The question: A mortgage is taken out for £80000. It is to be paid by annual instalments of £5000 with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.

HINT: Find an expression for the debt remaining after n years and solve using the fact that if it is paid off, the debt = 0.

What I started with: debt left = (80000 - 5000n)*1.04 - does not work
What I got up to was: After nth year debt left was = 1.04*(n-1 value - 5000)
Didn't not get anywhere with that either.
Thanks

2. Hello, person!

This confusion has turned up from time to time.

I don't know who is teaching that a time-payment plan is a simple geometric series.
It is very annoying and I wish they would stop!

A mortgage is taken out for $80,000. It is to be paid by annual instalments of$5000 with the first payment being made
at the end of the first year that the mortgage was taken out.
Interest of 4% is then charged on any outstanding debt.
Find the total time taken to pay off the mortgage.

HINT: Find an expression for the debt remaining after n years
and solve using the fact that if it is paid off, the debt = 0.

This is an Amortization problem.
You can (1) derive the Amortization formula, or (2) memorize the formula.
. . Neither task is pleasant.

Let: . $\begin{array}{ccc}P &=& \text{Principal borrowed} \\ r &=& \text{Periodic interest rate} \\ n &=& \text{No. of periods} \\ A &=& \text{Periodic payment} \end{array}$

We borrow $P$ dollars at time zero.

In one year, they charge r% interest.
At the end of year-1, we owe: . $P(1+r)$ dollars.
Then we pay $A$ dollars.
. . Our balance is: . $P(1+r) - A$ dollars.

During year-2, they charge r% interest.
At the end of year-2, we owe: . $(1+r)\bigg[P(1+r) - A\bigg]$ dollars.
Then we pay $A$ dollars.
. . Our balance is: . $(1+r)^2P - (1+r)A - A$ dollars.

During year-3, they charge r% interest.
At the end of year-3, we owe: . $(1+r)\bigg[(1+r)^2P - (1+r)A - A\bigg]$ dollars.
Then we pay $A$ dollars.
. . Our balance is: . $(1+r)^3P - (1+r)^2A - (1+r)A - A$ dollars.

See the pattern?

At the end of year- $n$, our balance is:
. . $(1+r)^nP - (1+r)^{n-1}A - (1+r)^{n-2}A - \hdots - A$ dollars.

But by year- $n$, we expect to pay off the loan; the balance is \$0.

So, we have: . $(1+r)^nP -(1+r)^{n-1}A - (1+r)^{n-2} - \hdots - A \;=\;0$

. . . . $(1+r)^nP \;=\;A\underbrace{\bigg[1 + (1+r)^2 + (1+r)^3 + \hdots + (1+r)^{n-1}\bigg]}_{\text{geomtric series}}$ .[1]

The geometric series has the sum: . $S \;=\;\frac{(1+r)^n - 1}{(1+r)-1} \;=\;\frac{(1+r)^n-1}{r}$

Then [1] becomes: . $(1+r)^nP \;=\;A\cdot\frac{(1+r)^n-1}{r}$

Therefore: . $\boxed{A \;=\;P\,\frac{r(1+r)^n}{(1+r)^n-1}}$ . . . the Amortization Formula !

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the problem . . .

We are given: . $P = 80,\!000,\;A = 5,\!000,\;r = 4\% = 0.04$

Substitute into the Formula: . $5,\!000 \;=\;80,\!000\,\frac{(0.04)(1.04^n)}{1.04^n-1}$

. . . . $1.04^n-1 \;=\;0.64(1.04^n) \quad\Rightarrow\quad 1.04^n -0.64(1.04^n) \;=\;1$

. . . . $0.36(1.04^n) \;=\;1\quad\Rightarrow\quad 1.04^n \;=\;\tfrac{1}{0.36}$

Take logs: . $\ln\left(1.04^n\right) \;=\; \ln\left(\tfrac{1}{0.36}\right) \quad\Rightarrow\quad n\ln(1.04) \;=\;\ln(\tfrac{1}{0.36})$

Therefore: . $n \;=\;\frac{\ln(\frac{1}{0.36})}{\ln(1.04)} \;=\;26.04876774 \;\approx\; 26$ years.

I need a nap . . .
.

3. Thank you!!!!!! That was longer than I had expected and much harder than thought I have not learnt that formula lol, take your nap and have some rep!