Geometric Series for Time Payments

Geometric Series for Time Payments

Time payments are calculated based on the 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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