A marathon is 26.2 miles. What is the least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon?

15 times

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