Geometric progression (compound interest)

Question

"A man, who started work in 1990, planned an investment for his retirement in 2030 in the following way. On the first day of each year, from 1990 to 2029 inclusive, he is to place £100 in an investment account. The account pays 10% compound interest per annum, and interest is added on the 31st December of each year of the investment. Calculate the value of his investment on 1st January 2030."

I understand that this follows a GP. At the end of 1990, he has $£100(1.1)$, at the end of 1991 he has $£100(1.1)(1.1)=£100(1.1)^2$ and so on till the end of 2029.

I managed to get the answer but wasn't satisfied with my method. Firstly, how do you work out the number of years inclusive, between a particular year and another, at least in an intuitive manner? Also with something like how many days are between 2 dates, or how many terms there are between a part of a series?

Also how can I do this in a formulated manner? I had to do this written out with end year 1, end year 2 etc. The regular GP formula doesn't work and I can appreciate why, sort of. Could a formula be used from the start of year 1?

Cheers

Answer 1

Here we want the value of the investment on 31st December 2029, which will be the same as the value on the 1st Jan 2030

So you start with...

On 31st Dec 1990, the value of his investment is $100\times 1.1$

On 31st Dec 1991 the value is $(100\times1.1+100)\times 1.1$

On 31st Dec 1992 the value is $((100\times1.1+100)\times1.1+100)\times 1.1$

This is the same as $100\times1.1(1+1.1+1.1^2)$

Clearly this is a GP with three terms. We can infer that

On 31st Dec 2029 the value is $100\times1.1(1+1.1+1.1^2+...+1.1^{39})$ which is a GP with $40$ terms, so the total is $$100\times1.1\times\frac{1.1^{40}-1}{1.1-1}$$

Note that we can count the number of £100 payments by doing $2029-1990+1$.

Answer 2

Let $P$ the periodic payment (deposit), $n$ the number of years over which payments are made and $r$ the annual interest rate.

The payment in the first year at the end of $n$ years will produce the value $S_1=P(1+r)^n$.

The payment in the second year at the end of $n-1$ years will produce the value $S_{2}=P(1+r)^{n-1}$. And so on.

The payment in the $n$-th year at the end of $n-(n-1)=1$ years will produce the value $S_n=P(1+r)^{n-(n-1)}=P(1+r)$.

So you have to sum all the future values produced of each payments: $$ S=S_{1}+S_{2}+\cdots+S_{n}=P(1+r)^{n}+P(1+r)^{n-1}+\cdots+P(1+r)=\sum_{t=1}^n P(1+r)^{t} $$ The sum is a geometric series, so we have $$ S=P(1+r)\frac{(1+r)^n-1}{r} $$ In financial mathematics, this result is the Future Value of Annuity Due.

In your case $P=100$, $r=10\%$ and $n=40$ (the 2030 is not included) so that $S=48,685.18$