# 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.

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\times 1.05^{1} &\color{green}{\text{compounded 1 year}} \\ 21^{\text{st}} \ \$100 &\leadsto \$100 &\color{green}{\text{withdrawn immediately after deposit}} \end{align} \) \( \begin{align} \require{AMSsymbols} \text{Total amount} &= \$100 + \$100 \times 10.5^1 + \$100 \times 1.05^2 +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

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