So the ordered pairs given tell us how much time has passed since the beginning at each station. If we analyze this data as a sequence we have that the terms of the sequence are given by the time and their orders are given by the station number. Then the terms of the sequence are:
[tex]4,8,16,32\ldots[/tex]As you can see the difference between consecutive terms is not always the same. This means that this is not an arithmetic sequence. So this is probably a geometric sequence. The general formula for such type of sequence is:
[tex]a_n=a_1\cdot r^{n-1}[/tex]Where the n denotes the term order, a₁ is the first term (in this case it's 4) and r is known as the common ratio. So we know that the first term is 4 so we get:
[tex]a_n=4\cdot r^{n-1}[/tex]We can use the other terms of the sequence to find r:
[tex]\begin{gathered} 8=a_2=4\cdot r^{2-1}=4\cdot r \\ 8=4r \end{gathered}[/tex]If we divide both sides of this equation by 4 we get the value of r:
[tex]\begin{gathered} \frac{8}{4}=\frac{4r}{4} \\ 2=r \end{gathered}[/tex]So r=2 and the general formula looks like this:
[tex]a_n=4\cdot2^{n-1}[/tex]If we take n=3 and n=4 we should get 16 and 32. Let's check this:
[tex]\begin{gathered} a_3=4\cdot2^{3-1}=4\cdot4=16 \\ a_4=4\cdot2^{4-1}=4\cdot8=32 \end{gathered}[/tex]So r is in deed equal to 2. Then the answer to part A is that this is a geometric sequence.
In part B we have to find the term for n=5. However we have to use a recursive formula and the one I just found is explicit. A recursive formula gives you the value of a term using the value of a previous term. Now let's see the differences between consecutive terms:
[tex]\begin{gathered} a_2-a_1=8-4=4=2^2 \\ a_3-a_2=16-8=8=2^3 \\ a_4-a_3=32-16=16=2^4 \end{gathered}[/tex]So we have that the difference between consecutive terms is given by:
[tex]a_n-a_{n-1}=2^n[/tex]Then the recursive formula we have to use is given by this expression:
[tex]a_n=a_{n-1}+2^n[/tex]Then the time when Aurora completes station 5 is given by:
[tex]a_5=a_{5-1}+2^5=a_4+32=32+32=64[/tex]Then the answer to part B is 64 minutes.
For part C we just need to use the explicit formula we found in part A:
[tex]a_n=4\cdot2^{n-1}[/tex]We have to take n=8:
[tex]a_8=4\cdot2^{8-1}=4\cdot2^7=4\cdot128=512[/tex]Then the answer to part C is 512 minutes.
