part A. We are given that a hiker will increase the distance covered by 10% each day. Let "S" be the distance, then on the first day the distance is:
[tex]S_1[/tex]On the second day, we must add 10% of the first day, we get:
[tex]S_1=S_1+\frac{10}{100}S_1[/tex]Simplifying we get:
[tex]S_2=S_1+0.1S_1=1.1S_1[/tex]On the third day, we add 10% of the second day, we get:
[tex]S_3=S_2+0.1S_2=1.1S_2=(1.1)(1.1)S_1=(1.1)^2S_1[/tex]On the fourth day, we add 10% of the third day, we get:
[tex]S_4=S_3+0.1S_3=1.1S_3=(1.1)^3S_1[/tex]If we continue this pattern and we set "n" as the number of days, then a formula for the distance after "n" days is:
[tex]S_n=(1.1)^{n-1}S_1[/tex]Now, we are given that for n = 7 the distance is 75897, therefore, we substitute n = 7 in the formula:
[tex]S_7=(1.1)^{7-1}S_1[/tex]Substituting the value of the distance:
[tex]75897=(1.1)^{7-1}S_1[/tex]Now we can solve for S1, we do that by dividing both sides by 1.1 together with its
exponent:
[tex]\frac{75897}{(1.1)^{7-1}}=S_1[/tex]Now we solve the operations:
[tex]\frac{75897}{(1.1)^6}=S_1[/tex]Solving the operations:
[tex]42842=S_1[/tex]Therefore, the distance the first day was 42842 miles.
part B. The formula for Sn is the given previously but we replace the known value of S1:
[tex]S_n=42842(1.1)^{n-1}[/tex]Part C. To determine the distance after 10 days, we substitute the value n = 10 in the formula, we get:
[tex]S_{10}=42842(1.1)^{10-1}[/tex]Solving the operations we get:
[tex]S_{10}=101019.19[/tex]Therefore, the distance after 10 days is 101019.19 miles.
