Geometric Series for Time Payments


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

Worked Examples of Geometric Series for Time Payments 1

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

(a)    Find the amount owing at the end of the first 3 months.

\( \begin{aligned} \displaystyle \require{red}
\text{after 1 month } \$100000 \times 1.01 – \$1200 &= \$99800 \\
\text{after 2 months } \$99800 \times 1.01 – \$1200 &= \$99589 \\
\therefore \text{after 3 months } \$99589 \times 1.01 – \$1200 &= \$99393.98 \\
\end{aligned} \\ \)

(b)    Find the amount owing at the end of \(n\) months.

\( \begin{aligned} \displaystyle \require{green}
\text{1M: } &\$10000 \times 1.01 – \$1200 \\
\text{2M: } &(\$100000 \times 1.01 – \$1200) \times 1.01 – \$1200 \\
=&\$100000 \times 1.01^2 – \$1200 \times 1.01 – \$1200 \\
=&\$100000 \times 1.01^2 – \$1200 \times (1.01 + 1) \\
\text{3M: } &(\$100000 \times 1.01^2 – \$1200 \times 1.01 – \$1200 ) \times 1.01 – \$1200 \\
=&\$100000 \times 1.01^3 – \$1200 \times 1.01^2 – \$1200 \times 1.01 – \$1200 \\
=&\$100000 \times 1.01^3 – \$1200 \times (1.01^2 + 1.01 + 1) \\
n \text{M: } &$100000 \times 1.01^n – \$1200 \times (1.01^{n-1} + 1.01^{n-2} + \cdots + 1.01 + 1 ) \\
=&\$100000 \times 1.01 ^n – \$1200 \times \frac{1.01^n-1}{1,01-1} \\
=&\$100000 \times 1.01^n – \$120000 \times (1.01^n -1) \\
=&\$100000 \times 1.01^n – \$120000 \times 1.01^n + \$120000 \\
=&\$120000 – \$20000 \times 1.01^n \\
\end{aligned} \\ \)

(c)    Find how long it takes to repay the full loan.

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

(d)    Find the amount owing at the end of n months if he is able to pay $2000 per month.

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

(e)    How long would repayment take if the monthly repayment is $2000?

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

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