Respuesta :
The question can be solved using an arithmetic sequence model.
The general formula for an arithmetic sequence is given to be:
[tex]a_n=a_1+(n-1)d[/tex]where
[tex]\begin{gathered} a_n=\text{ nth term of the sequence} \\ a_1=\text{ first term} \\ d=\text{ common difference} \\ n=\text{ number/position of nth term} \end{gathered}[/tex]We can get the common difference by applying the information given into the formula.
GIVEN:
[tex]\begin{gathered} a_0=30\text{ (Initial population)} \\ a_8=670 \end{gathered}[/tex]We can model the formula for the 8th term such that:
[tex]a_8=a_1+(8-1)d[/tex]We have the value for the first term to be:
[tex]a_1=a_0+d[/tex]Substituting the values given, we have:
[tex]\begin{gathered} 670=(a_0+d)+7d \\ 670=30+d+7d \\ 670=30+8d \\ 8d=670-30 \\ 8d=640 \\ d=\frac{640}{8} \\ d=80 \end{gathered}[/tex]QUESTION A:
The general recursive formula is given to be:
[tex]a_n=a_{n-1}+d[/tex]Therefore, for this sequence, the recursive formula is:
[tex]a_n=a_{n-1}+80[/tex]QUESTION B:
The general explicit formula for an arithmetic sequence is given to be:
[tex]a_n=a_1+(n-1)d[/tex]Therefore, for this sequence, the explicit formula is given to be:
[tex]\begin{gathered} a_n=a_0+d+(n-1)d \\ a_n=a_0+d+dn-d \\ a_n=a_0+dn \\ \text{Substituting values of }a_0\text{ and d, we have:} \\ a_n=30+80n \end{gathered}[/tex]QUESTION C:
In week 62, we will take n = 62:
[tex]\begin{gathered} a_{62}=30+80(62) \\ a_{62}=4990 \end{gathered}[/tex]There will be 4990 beetles in week 62.
QUESTION D:
In this problem, we will make the following substitution:
[tex]a_n=1000[/tex]Therefore, we can substitute into the explicit formula and solve for n as shown:
[tex]\begin{gathered} 1000=30+80n \\ 80n=1000-30 \\ 80n=970 \\ n=\frac{970}{80} \\ n=12.125 \end{gathered}[/tex]Therefore, in the 13th week, since we can't use a decimal for the weeks, the beetle population will reach 1000 beetles.