# Thread: geometric and arithmetic sequences

1. ## geometric and arithmetic sequences

On an island the population of a species of insect (species A) is increasing geometrically, with a population of 25,000 in 1990 & an annual growth rate of 13.5%. Another species of insect (species B) is increasing in population arithmetically with numbers 37,000 in 1990 & an annual increment of 3500.

1. determine the difference in the numbers of the 2 species for each year during a decade (up to 1999)
2. In what year will species A be greater in number than species B? Assuming the growth rates remain fixed.

A scientist has a mathematical model where the species can cohabit provided that they have equal numbers in the year 2000.

3. If the growth rate in species A is to remain unchanged, what would the annual increment in species B need to be to achieve this?
4. If the annual increment in species B is to remain unchanged, what would the growth rate in species A need to be to achieve this?

2. 1. You should know for an arithmetic sequence that

$\displaystyle t_n = t_0 + nd$

and for a geometric sequence that

$\displaystyle t_n = t_0r^n$.

So for species $\displaystyle A$, the population during year $\displaystyle n$ is

$\displaystyle A_n = 25\,000\cdot 0.135^n$

and for species $\displaystyle B$, the population during year $\displaystyle n$ is

$\displaystyle B_n = 37\,000 + 3500n$.

2. You want species $\displaystyle A$ to be greater in number than species $\displaystyle B$, so

$\displaystyle 25\,000\cdot 0.135^n > 37\,000 + 3500n$.

You will need to use technology to solve this inequality for $\displaystyle n$.