# 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}