Missing part of the question
Miguel has started training for a race. The first time he trains, he runs 0.5 mile. Each subsequent time he trains, he runs 0.2 mile farther than he did the previous time.
What is the arithmetic series that represents the total distance Miguel has run after he has trained n times?
Answer:
The least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon is 17.3 miles
Step-by-step explanation:.
Given parameters
Miguel first run = 0.5 mile
Subsequent run = 0.2 mile
This question is an arithmetic progression.
We'll make use of arithmetic progression formula to solve this
Formula:.
Tn = a + (n - 1)d
Where a = first term
n = number of terms
d = common difference
In this case
a = first run = 0.5 mile
d = subsequent run = 0.2 mile
So, Tn = a + (n - 1)d become
Tn = 0.5 + (n - 1) 0.2
Tn = 0.5 + 0.2n - 0.2
Tn = 0.5 - 0.2 + 0.2n
Tn = 0.3 + 0.2n
The arithmetic series of an arithmetic progression is calculated using
Sn = ½(a + Tn) * n
By substituton, we have
Sn = ½(0.5 + 0.3 + 0.2n) * n
Sn = ½(0.8 + 0.2n) * n
Sn = 0.4n + 0.1n²
b.
Since the race is 26.2 miles then the least number of times is given as
Sn ≥ 26
0.4n + 0.1n² ≥ 26.2
0.1n² + 0.4n - 26.2 ≥ 0
Using quadratic formula
n = (-b ± √(b² - 4ac))/2a
Where b = 0.4 a = 0.1 and C = -26.2
So,
n = -0.4 ± √(0.4² - 4 * 0.1 * ,26.2)/2 * 0.1
n = (-0.4 ± √10.64)/0.2
n = (0.4 ± 3.26)/0.2
n = (0.4 + 3.26)/0.2 or (0.4 - 3.26)/0.2
n = 3.46/0.2 or -2.86/0.2
n = 17.3 or -14.3
Since n can't be negative
n = 17.3 miles
The least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon is 17.3 miles
