c He will run 100 miles in total during the 9^(th) day.
Practice makes perfect
a Let a_n be the miles that Adrahan runs during the n^(th) day. On the first day he runs 2 miles. This tells us that a_1=2. He runs 1.5 times the distance he ran the time before. Therefore, a_n is a geometric sequence with the common ratio r=1.5.
a_n--- Geometric Sequence
a_1=2, r=1.5
We are asked to write the first five terms of a_n. Let's do it!
b We are asked to find the day when he will exceed 26 miles in one run. Let's find the few next terms of this sequence. From Part B we know that a_5=10.125. Each consecutive terms is 1.5 times the previous one.
n
a_n=1.5a_(n-1)
a_n
6
a_6=1.5a_5=1.5(10.125)
a_6≈ 15.19
7
a_7=1.5a_6=1.5(15.19)
a_7≈ 22.79
8
a_8=1.5a_7=1.5(22.79)
a_8≈ 32.19
Therefore, Adrahan will exceed 26 miles during the 8^(th) day.
c We are asked to find the day when he will run 100 miles in total. First, let S_n denote the sum a_1+a_2+... +a_n. To find S_n, we will use the formula for the sum of a geometric series.
S_n=a_1(1- r^n)/1- rFrom Part A we know that a_1= 2 and r= 1.5. Let's substitute the known values into the formula.
Next, we want to find the smallest value of n such that S_n=4(1.5)^n-4 will exceed 100. Let's do it!
n
4(1.5)^n-4
S_n=4(1.5)^n-4
1
4(1.5)^1-4
S_1=2
2
4(1.5)^2-4
S_2=5
3
4(1.5)^3-4
S_3=9.5
4
4(1.5)^4-4
S_4=16.25
5
4(1.5)^5-4
S_5≈ 26.38
6
4(1.5)^6-4
S_6≈ 41.56
7
4(1.5)^7-4
S_7≈ 64.34
8
4(1.5)^8-4
S_8≈ 98.52
9
4(1.5)^9-4
S_9≈ 149.77
Therefore, Adrahan will run 100 miles in total during the 9^(th) day.
Extra
Calculations Using a Spreadsheet
We will use a spreadsheet to find the smallest value of n such that S_n=4(1.5)^n-4 will exceed 100. Let's enter 1 in cell A1 and write =A1+1 in cell A2. When we hit enter we will get the following.
Next, copy cell A2, highlight cells A3 through A11, and paste.
Now, write =Round(4*Power(1.5,A1)-4,2) in cell B1. Next, copy cell B1, highlight cells B2 through B11, and paste.
Therefore, Adrahan will run 100 miles in total during the 9^(th) day.
