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Answer :
Sure, let's break down the steps for solving this athlete's training problem:
### Part a: How far does the athlete run on the 15th day?
The athlete increases the running distance each day by a consistent amount. This is an arithmetic sequence problem where:
- The distance on the first day is 600 meters (this is the first term of the sequence).
- The daily increment in running distance is 300 meters (common difference).
To find the distance on the 15th day, we can use the formula for the [tex]\(n\)[/tex]th term of an arithmetic sequence:
[tex]\[
a_n = a_1 + (n-1) \times d
\][/tex]
where:
- [tex]\(a_n\)[/tex] is the distance on the nth day,
- [tex]\(a_1 = 600\)[/tex] meters is the distance on the first day,
- [tex]\(d = 300\)[/tex] meters is the common difference,
- [tex]\(n = 15\)[/tex].
Plugging in the values:
[tex]\[
a_{15} = 600 + (15-1) \times 300
\][/tex]
[tex]\[
a_{15} = 600 + 14 \times 300
\][/tex]
[tex]\[
a_{15} = 600 + 4200
\][/tex]
[tex]\[
a_{15} = 4800 \text{ meters}
\][/tex]
So, on the 15th day, the athlete runs 4800 meters.
### Part b: What's the total distance that he will run in 15 days?
To find the total distance run over 15 days, we use the formula for the sum of the first [tex]\(n\)[/tex] terms of an arithmetic sequence:
[tex]\[
S_n = \frac{n}{2} \times (2a_1 + (n-1) \times d)
\][/tex]
where:
- [tex]\(S_n\)[/tex] is the total distance run in [tex]\(n\)[/tex] days,
- [tex]\(n = 15\)[/tex],
- [tex]\(a_1 = 600\)[/tex],
- [tex]\(d = 300\)[/tex].
Plugging in the values:
[tex]\[
S_{15} = \frac{15}{2} \times (2 \times 600 + (15-1) \times 300)
\][/tex]
[tex]\[
S_{15} = \frac{15}{2} \times (1200 + 4200)
\][/tex]
[tex]\[
S_{15} = \frac{15}{2} \times 5400
\][/tex]
[tex]\[
S_{15} = 15 \times 2700
\][/tex]
[tex]\[
S_{15} = 40500 \text{ meters}
\][/tex]
So, the total distance run over 15 days is 40,500 meters.
### Part c: How long will it be before he can run a marathon of 42 km?
We need to determine the day on which the cumulative distance will first equal or exceed 42,000 meters (since 1 kilometer = 1000 meters).
Using the formula for the sum of an arithmetic sequence [tex]\(S_n \geq 42000\)[/tex]:
[tex]\[
S_n = \frac{n}{2} \times (2a_1 + (n-1) \times d)
\][/tex]
Solving for [tex]\(n\)[/tex] when [tex]\(S_n \geq 42000\)[/tex]:
The calculations show that it will be on the 139th day that the total running distance first reaches or exceeds 42,000 meters.
Therefore, it will take 139 days until the athlete can run a total of 42 km.
### Part a: How far does the athlete run on the 15th day?
The athlete increases the running distance each day by a consistent amount. This is an arithmetic sequence problem where:
- The distance on the first day is 600 meters (this is the first term of the sequence).
- The daily increment in running distance is 300 meters (common difference).
To find the distance on the 15th day, we can use the formula for the [tex]\(n\)[/tex]th term of an arithmetic sequence:
[tex]\[
a_n = a_1 + (n-1) \times d
\][/tex]
where:
- [tex]\(a_n\)[/tex] is the distance on the nth day,
- [tex]\(a_1 = 600\)[/tex] meters is the distance on the first day,
- [tex]\(d = 300\)[/tex] meters is the common difference,
- [tex]\(n = 15\)[/tex].
Plugging in the values:
[tex]\[
a_{15} = 600 + (15-1) \times 300
\][/tex]
[tex]\[
a_{15} = 600 + 14 \times 300
\][/tex]
[tex]\[
a_{15} = 600 + 4200
\][/tex]
[tex]\[
a_{15} = 4800 \text{ meters}
\][/tex]
So, on the 15th day, the athlete runs 4800 meters.
### Part b: What's the total distance that he will run in 15 days?
To find the total distance run over 15 days, we use the formula for the sum of the first [tex]\(n\)[/tex] terms of an arithmetic sequence:
[tex]\[
S_n = \frac{n}{2} \times (2a_1 + (n-1) \times d)
\][/tex]
where:
- [tex]\(S_n\)[/tex] is the total distance run in [tex]\(n\)[/tex] days,
- [tex]\(n = 15\)[/tex],
- [tex]\(a_1 = 600\)[/tex],
- [tex]\(d = 300\)[/tex].
Plugging in the values:
[tex]\[
S_{15} = \frac{15}{2} \times (2 \times 600 + (15-1) \times 300)
\][/tex]
[tex]\[
S_{15} = \frac{15}{2} \times (1200 + 4200)
\][/tex]
[tex]\[
S_{15} = \frac{15}{2} \times 5400
\][/tex]
[tex]\[
S_{15} = 15 \times 2700
\][/tex]
[tex]\[
S_{15} = 40500 \text{ meters}
\][/tex]
So, the total distance run over 15 days is 40,500 meters.
### Part c: How long will it be before he can run a marathon of 42 km?
We need to determine the day on which the cumulative distance will first equal or exceed 42,000 meters (since 1 kilometer = 1000 meters).
Using the formula for the sum of an arithmetic sequence [tex]\(S_n \geq 42000\)[/tex]:
[tex]\[
S_n = \frac{n}{2} \times (2a_1 + (n-1) \times d)
\][/tex]
Solving for [tex]\(n\)[/tex] when [tex]\(S_n \geq 42000\)[/tex]:
The calculations show that it will be on the 139th day that the total running distance first reaches or exceeds 42,000 meters.
Therefore, it will take 139 days until the athlete can run a total of 42 km.
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