Geometric Sequence Problems

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

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

Geometric Sequence Problems Formula

Example 1

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

(a)   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)   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 in an island at the beginning of $2018$ was $560$. The population has been steadily decreasing by $4$% per year.

(a)   Find the population at the beginning of 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 population of rabbits will be approx 495 at the beginning of $2022$.

(b)   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 population of rabbits will be approx $1405$ at $35$ years’ time.
So it will be $2053$.

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