# Geometric Series for Time Payments

Time payments are calculated based on Geometric Series for reducible compound interests. Basically, the 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 is2000?

\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} \\