Answer:
This is a simple Arithmetic Progression question, and we will use the formula:[tex]T_{n}= a_{1} + (n-1)d\\[/tex]
Step-by-step explanation:
From the question, we started with 1,000 mosquitoes, this is the first term ([tex]a_{1}[/tex]).
The next term is 1,200.
This means that the Common Difference ([tex]d[/tex]) is:
1,200 - 1,000 = 200.
The [tex]nth[/tex] term ([tex]n[/tex]) we are looking for is 22.
So we have:
[tex]a_{1}[/tex] = 1,000
[tex]n[/tex] = 22
[tex]d[/tex] = 200
We solve as follows:
[tex]T_{n}= a_{1} + (n-1)d\\[/tex]
[tex]T_{22}[/tex] = 1,000 + (22-1)200
[tex]T_{22}[/tex] = 1,000 + (21)200
[tex]T_{22}[/tex] = 1,000 + 4,200
[tex]T_{22}[/tex] = 5,200.
So therefore, after 22 days, the number of mosquitoes in the colony will be 5,200 mosquitoes.
For the other question, we will use the value 40,000 mosquitoes.
We solve thus;
since it took 22 days to get to 5,200 mosquitoes, it will therefore take [tex]x[/tex] days to get 40,000 mosquitoes
22 days = 5,200 mosquitoes
[tex]x[/tex] days = 40,000 mosquitoes
we cross multiply thus:
22 x 40,000 = 5,200 x [tex]x[/tex]
880,000 = 5,200[tex]x[/tex]
making [tex]x[/tex] the subject, we have:
[tex]x[/tex] = [tex]\frac{880,000}{5,200}[/tex]
[tex]x[/tex] = 169
Therefore, it will take 169 days to get 40,000 mosquitoes.
