# Geometric Sequence Problems

Growth and decay problems involve repeated multiplications by a constant number, a common ratio. We can thus use geometric sequences to model these situations.

\large \begin{align} \require{AMSsymbols} \displaystyle \require{color} \color{red}u_{n} &= u_{1} \times r^{n-1} \\ \require{color} \color{red}u_{n+1} &= u_{1} \times r^{n} \end{align}

## Example 1

The initial population of chickens on a farm was $40$. The population increased by $5$% each week.

(a) â€ƒ How many chickens were present after $20$ weeks?

\begin{align} \displaystyle u_{1} &= 40 \\ r &= 1 + 0.05 = 1.05 \\ u_{n} &= 40 \times 1.05^{n-1} \\ u_{20} &= 40 \times 1.05^{20-1} \\ &= 101.078 \cdots \end{align}
There were $101$ chickens.

(b) â€ƒ How long would it take for the population to reach $240$?

\begin{align} \displaystyle 40 \times 1.05^{n-1} &= 240 \\ 1.05^{n-1} &= 6 \\ n-1 &= \log_{1.05}{6} \\ n &= \log_{1.05}{6} + 1 \\ &= \dfrac{\log_{10}{6}}{\log_{10}{1.05}}\ + 1 \\ &= 37.723 \cdots \end{align}
The population will reach $240$ in the $38$th week.

## Example 2

The population of rabbits on an island at the beginning of $2018$ was $560$. The population has been steadily decreasing by $4$% per year.

(a) â€ƒ Find the population at the beginning of the year $2022$.

\begin{align} \displaystyle u_{1} &= 560 &2018\\ r &= 1-0.04 = 0.96 \\ u_{4} &= 560 \times 0.96^3 &2022 \\ &= 495.45 \cdots \end{align}
The rabbit population will be approximately $495$ at the beginning of $2022$.

(b) â€ƒ In which year would we expect that population to have declined to $140$?

\begin{align} \displaystyle 560 \times 0.96^{n-1} &= 140 \\ 0.96^{n-1} &= 140 \div 560 \\ &= 0.25 \\ n-1 &= \log_{0.96}{0.25} \\ n &= \log_{0.96}{0.25} + 1 \\ &= \dfrac{\log_{10}{0.25}}{\log_{10}{0.96}} + 1 \\ &= 34.959 \cdots \\ \end{align}
The rabbits’ population will be approximately $1405$ at $35$ years.
So it will be $2053$.

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