# compound interest with geometric series

Were studying geometric sequences in maths and this came up as one of the questions:

A mortgage is taken out for 150000 and is repaid annually with 20000 installments. Interest is charged on the outstanding debt at 10%, calculated annually. If the first repayment us made one year after the mortgage is taken out, find the number of years it takes for the mortgage to be repaid.

When i looked up the solutions online to this, they didn't use a geometric series formula to solve they just did it manually and they eventually got to an answer of 15 years. My maths teacher told us to first find the Tn formula (ar the power of n-1) and then try and use that to form the series formula . I don't know how to write out the series formula here so can anyone help me ?

• Hopefully someone who knows finance stuff and understands what "outstanding debt" and "installments" are can help, though this seems to be an otherwise simple problem of exponential growth-decay (compound interest and stuff)... Dec 3, 2013 at 21:28
• thanks for answering .. by outstanding debt i mean you add on 10% percent after taking 20000 away from the current amount. for example at the end of yr 1 the amount to be paid is 155000(it says 20000 doesn't begin to be paid until the second year) in yr 2 20000 is taken away from 155000 and then 10% is added on for interest.
– Zach
Dec 3, 2013 at 21:43
• by installments i meant every year he/she pays 20000 of the debt + 10% every year until its paid off.
– Zach
Dec 3, 2013 at 21:45
• About your first comment: after the first year the ammount owed is $\;150,000+150,000\cdot 0.1=165,0004,$ and then minus $\;20,000\;$ paid that'd equal $145,000\;$. Next year the ammount owed is $\;145,000+0.1\cdot145,000=159,500\;$ and then minus $\;20,000\;$ we get a debth of $\;139,500\;$ , etc. If this is what you meant this comment also hints you how you can deduce by yourself what to do... Dec 3, 2013 at 21:48
• sorry, ive got it completely wrong its add interst first then minus 20000. im a bit tired didn't realise what i was writing. In the solution it says yr 2 was 150000*1.1 = 165000 - 20000 = 145000 (excluding year one since that came up to 165000 anyway since the person didn't start paying the 20000 until the second year) this 145000 goes through the same process again and eventually you get to an answer of 15 years. I was just wondering what is the geometric series formula for this .
– Zach
Dec 3, 2013 at 21:56

There are two ways you can try to solve this problem. The first is try try to keep a running tab of how much is still owed at time $n$. For the sake of legibility, I am going to divide the amounts by 1000. The recursive formula you get is then:

$$a_0=150,\qquad a_{n+1}=(1+0.1)a_n-20.$$

If you compute the first few terms of this sequence, you get

$$a_1=150(1.1)-20, a_2=150(1.1)^2-20(1.1)-20, a_3=150(1.1)^3-20(1.1)^2-20(1.1)-20\ldots$$

We can factor and get a general formula

$$a_n=150(1.1)^n-20(1+1.1+1.1^2+\ldots+1.1^{n-1})$$

We have that $a_n$ is the difference of two terms: one is a geometric series, and the other is growing exponentially. We can use the formula for the sum of a geometric series to get

$$a_n=150(1.1)^n-20\frac{(1.1)^n-1}{0.1}.$$

However, we can get to this formula more quickly by reinterpreting what interest means. If you have investment opportunities, a dollar today is worth more than a dollar tomorrow, because you can invest that dollar to earn interest. If an investment of $1$ dollar today yields $1+i$ dollars tomorrow, then $1$ dollar tomorrow is worth only $1/(1+i)$ dollars today. We can then treat interest calculations as being akin to currency conversion calculations. The same way you might convert a dollar into some number of euros, when interest is computed you are converting today dollars into tomorrow dollars (or next year dollars, or whatever your interest periods are).

Now, we want to reinterpret the problem by converting all future dollar amounts into today-dollars. This has the advantage that the $150$ in our formulas never changes. It is just our initial amount measured in today dollars. However, the amount paid in future dollars has to be converted. Let $b_n$ be the amount still owed, but this time measured in today-dollars. The amount we pay back in year $n$ is then $20(1.1)^{-n}$, and so

$$b_n=150-20(1.1)^{-1}-20(1.1)^{-2}-\ldots -20(1.1)^{-n}=150-20(1.1)^{-1}\left(\frac{1-(1.1)^{-n}}{1-(1.1)^{-1}} \right).$$

If you clean up the expression slightly, you can verify that $a_n=b_n(1.1)^n$, which is what we would hope: to calculate the amount we owe at time $n$, we can do all our calculations in terms of today-dollars, and then convert today-dollars into time-$n$ dollars.

Regardless, we want to solve for the $n$ that makes $b_n=0$. We get

$$1-(1.1)^{-n}=3/4 \Longrightarrow (1.1)^{-n}=1/4 \Longrightarrow n=\log 4/\log 1.1\approx 14.55$$

However, this isn't a whole number, so we have to be careful in interpreting this "answer." Since we only pay back the loan once a year, the result tells us that we are not done after the $14$th year, but we are done by the $15$th year. Our final payment in the $15$th year will be less than the $20$ we pay every other year. To find out how much we need to reduce our final payment by, let us calculate $a_{15}\approx -8.86$. Thus, if we did pay 20 in the final year, we would have overpaid by $8.86$, and thus our final payment should be for $20-8.86=11.14$.

You're trying to solve the equation $150000 = \displaystyle Pa_{\overline{n}|i} = P\frac{1-(1+i)^{-n}}{i}$ for $n$, where $a_{\overline{n}|i}$ is a financial annuity (basically, a fancy name finance people use for a finite geometric series). Plugging in values,

$150000 = 20000 \frac{1-(1.1)^{-n}}{0.1} \implies 1.1^{-n}=0.25 \implies n \approx 14.545$.

The $NPV$ (net present value of future cashflows) $= \dfrac{CF}{(1+y)^t}$

Where CF is the cash flow, at t is the time when it is received.

The ballance of the mortgage equals the NPV of all of the future cashlows.

$150,000 = \sum_\limits{i=1}^n\dfrac{20,000}{(1.10)^i}\\ 150,000 = 20,000 \dfrac{1-\frac{1}{1.10}^n}{0.10}\\$

And solve for n.

n isn't going to be an integer. So there will be [n] payments of 20,000 and 1 payment of something less than that.

• After writing this response, I see that the question is old. Any reason why it is at the top of the questions queue? May 4, 2016 at 21:46