# Geometric Series for Time Payments

Time payments are calculated based on Geometric Series for reducible compound interests. The geometric series formula is used for handling this situation.

## Worked on Examples of Geometric Series for Time Payments 1

John took out a loan of $100 000$ on $1$st January $2001$ for home loan. Interest is charged at $12 \%$ per annum reducible, compounded monthly, and he will repay the loan in monthly instalments of $1200$.

## Part 1

Find the amount owing at the end of the first three months.

\begin{aligned} \displaystyle \text{after 1 month } \100000 \times 1.01-\1200 &= \99800 \\ \text{after 2 months } \99800 \times 1.01-\1200 &= \99589 \\ \therefore \text{after 3 months } \99589 \times 1.01-\1200 &= \99393.98 \end{aligned}

## Part 2

Find the amount owing at the end of $n$ months.

\begin{aligned} \displaystyle \text{1M: } &\10000 \times 1.01-\1200 \\ \text{2M: } &(\100000 \times 1.01-\1200) \times 1.01-\1200 \\ =&\100000 \times 1.01^2-\1200 \times 1.01-\1200 \\ =&\100000 \times 1.01^2-\1200 \times (1.01 + 1) \\ \text{3M: } &(\100000 \times 1.01^2-\1200 \times 1.01-\1200 ) \times 1.01-\1200 \\ =&\100000 \times 1.01^3-\1200 \times 1.01^2-\1200 \times 1.01-\1200 \\ =&\100000 \times 1.01^3-\1200 \times (1.01^2 + 1.01 + 1) \\ n \text{M: } &100000 \times 1.01^n-\1200 \times (1.01^{n-1} + 1.01^{n-2} + \cdots + 1.01 + 1 ) \\ =&\100000 \times 1.01 ^n-\1200 \times \frac{1.01^n-1}{1.01-1} \\ =&\100000 \times 1.01^n-\120000 \times (1.01^n-1) \\ =&\100000 \times 1.01^n-\120000 \times 1.01^n + \120000 \\ =&\120000-\20000 \times 1.01^n \end{aligned}

## Part 3

Find how long it takes to repay the full loan.

\begin{aligned} \displaystyle \120000-\20000 \times 1.01^n &= 0 \\ \20000 \times 1.01^n &= \120000 \\ 1.01^n &= 6 \\ n &= \log_{1.01} {6} \\ &= \frac{\log_{e}{6}}{\log_{e}{1.01}} \\ &= 180.08 \cdots \\ \therefore n &= 15 \text{ years 1 month} \end{aligned}

## Part 4

Find the amount owing at the end of n months if he can pay \$2000 per month.

\begin{aligned} \displaystyle &\100000 \times 1.01^n-\1200 \times (1.01^{n-1} + 1.01^{n-2} + \cdots + 1.01 + 1) \\ =&\100000 \times 1.01^n-\1200 \times \frac{1.01^n-1}{1.01-1} \\ =&200000-\100000 \times 1.01^n \end{aligned}

## Part 5

How long would repayment take if the monthly repayment is $\2000$?

\begin{aligned} \displaystyle \200000-\100000 \times 1.01^n &= 0 \\ 1.01^n &= 2 \\ n &= \log_{1.01}{2} \\ &= \frac{\log_{e}{2}}{\log_{e}{1.01}} \\ &= 69.6607 \cdots \\ \text{This is 70 months, that is } &\text{5 years and 10 months.} \end{aligned}

## Worked on Examples of Geometric Series for Time Payments 2

Amy’s parents have put $\100$ into a bank account each year on her birthday, starting with her first birthday. The account earns $5\%$ annually. Amy plans to withdraw all the money in the bank account on her twenty-first birthday, immediately after her parents make the twenty-first deposit.

## Part 1

If Amy withdraws all of the money in the account on her twenty-first birthday, how much of the total amount came from the $\100$ deposited in her account on her first birthday?

The number of years between the first and the twenty-first birthdays is $20$.
\begin{align} \100 \times 1.05^{20} &= \265.32977 \cdots \\ &\approx \265.33 \end{align}

## Part 2

Amy is wondering how much money she will have on her twenty-first birthday since each $\100$ deposit will have been earning interest for a different length of time.

\begin{align} \require{AMSsymbols} \displaystyle 1^{\text{st}} \ \100 &\leadsto \100 \times 1.05^{20} &\color{green}{\text{compounded 20 years}} \\ 2^{\text{nd}} \ \100 &\leadsto \100 \times 1.05^{19} &\color{green}{\text{compounded 19 years}} \\ 3^{\text{rd}} \ \100 &\leadsto \100 \times 1.05^{18} &\color{green}{\text{compounded 18 years}} \\ &\cdots \\ 19^{\text{th}} \ \100 &\leadsto \100 \times 1.05^{2} &\color{green}{\text{compounded 2 years}} \\ 20^{\text{th}} \ \100 &\leadsto \100 \times 1.05^{1} &\color{green}{\text{compounded 1 year}} \\ 21^{\text{st}} \ \100 &\leadsto \100 &\color{green}{\text{withdrawn immediately after deposit}} \end{align} \begin{align} \require{AMSsymbols} \text{Total amount} &= \100 + \100 \times 10.5^1 + \100 \times 1.05^2 + \cdots + \100 \times 1.05^{18} + \100 \times 1.05^{19} + \100 \times 1.05^{20} \\ &= \100 \left( 1+1.05^1 + 1.05^2 + \cdots + 1.05^{18} + 1.05^{19} + 1.05^{20} \right) \ \ \ \color{green}{\text{21 terms, first term 1, common ratio 1.05}} \\ &= \displaystyle \100 \times \frac{1(1.05^{21}-1)}{1.05-1} \ \ \ \color{green}{S_{n} = \frac{a(r^n-1)}{r-1}} \\ &= \3571.9251 \cdots \\ &\approx \3571.93 \end{align}