Solve/answer the following (show your solution in each problem)

1. One day, Isabela Jean jogged ½ kilometer. Because it was fun, she decided to increase the jogging distance each day y a certain amount. If she jogged 6 kilometer on the 51st day, how much more the distance increased each day?

2. Logs are stacked in a pile. The top row has 15 logs and the bottom has 21 logs. How many logs are in stack?

3. Insert five arithmetic means between 5 and 15.

### 1. **One day, Isabela Jean jogged ½ kilometer. Because it was fun, she decided to increase the jogging distance each day by a certain amount. If she jogged 6 kilometers on the 51st day, how much more did the distance increase each day?**

We are dealing with an **arithmetic sequence** here, where:
- The first term \( a_1 = \frac{1}{2} \) km.
- The 51st term \( a_{51} = 6 \) km.

The general formula for the \( n \)-th term of an arithmetic sequence is:
\[
a_n = a_1 + (n - 1) \cdot d
\]
where \( d \) is the common difference (the amount the distance increases each day).

We know:
- \( a_1 = \frac{1}{2} \),
- \( a_{51} = 6 \),
- \( n = 51 \).

Substituting into the formula for \( a_{51} \):
\[
6 = \frac{1}{2} + (51 - 1) \cdot d
\]
\[
6 = \frac{1}{2} + 50d
\]
Subtract \( \frac{1}{2} \) from both sides:
\[
6 - \frac{1}{2} = 50d
\]
\[
\frac{12}{2} - \frac{1}{2} = 50d
\]
\[
\frac{11}{2} = 50d
\]
Multiply both sides by 2:
\[
11 = 100d
\]
Solve for \( d \):
\[
d = \frac{11}{100} = 0.11
\]
Thus, the distance increased by **0.11 kilometers each day**.

### 2. **Logs are stacked in a pile. The top row has 15 logs and the bottom has 21 logs. How many logs are in the stack?**

The logs form an arithmetic sequence where:
- The top row (first term) \( a_1 = 15 \),
- The bottom row (last term) \( a_n = 21 \),
- The number of rows \( n \) is not given, but we know the difference \( d = 1 \) (since the number of logs decreases by 1 per row).

We use the formula for the sum of an arithmetic sequence:
\[
S_n = \frac{n}{2} \cdot (a_1 + a_n)
\]
We need to find \( n \). Using the formula for the \( n \)-th term:
\[
a_n = a_1 + (n - 1) \cdot d
\]
Substitute values:
\[
21 = 15 + (n - 1) \cdot 1
\]
\[
21 = 15 + n - 1
\]
\[
21 = n + 14
\]
Solve for \( n \):
\[
n = 21 - 14 = 7
\]
So, there are 7 rows of logs. Now, calculate the total number of logs using the sum formula:
\[
S_7 = \frac{7}{2} \cdot (15 + 21)
\]
\[
S_7 = \frac{7}{2} \cdot 36
\]
\[
S_7 = 7 \cdot 18 = 126
\]
Thus, there are **126 logs** in the stack.

### 3. **Insert five arithmetic means between 5 and 15.**

We need to insert five arithmetic means between the numbers 5 and 15. This means we are creating an arithmetic sequence with 7 terms, where:
- \( a_1 = 5 \),
- \( a_7 = 15 \).

The general formula for the \( n \)-th term of an arithmetic sequence is:
\[
a_n = a_1 + (n - 1) \cdot d
\]
We have:
\[
15 = 5 + (7 - 1) \cdot d
\]
\[
15 = 5 + 6d
\]
Subtract 5 from both sides:
\[
10 = 6d
\]
Solve for \( d \):
\[
d = \frac{10}{6} = \frac{5}{3}
\]

Now, we can find the five arithmetic means by adding \( d = \frac{5}{3} \) to each successive term:
- \( a_2 = 5 + \frac{5}{3} = \frac{15}{3} + \frac{5}{3} = \frac{20}{3} \approx 6.67 \),
- \( a_3 = a_2 + \frac{5}{3} = \frac{20}{3} + \frac{5}{3} = \frac{25}{3} \approx 8.33 \),
- \( a_4 = a_3 + \frac{5}{3} = \frac{25}{3} + \frac{5}{3} = \frac{30}{3} = 10 \),
- \( a_5 = a_4 + \frac{5}{3} = 10 + \frac{5}{3} = \frac{30}{3} + \frac{5}{3} = \frac{35}{3} \approx 11.67 \),
- \( a_6 = a_5 + \frac{5}{3} = \frac{35}{3} + \frac{5}{3} = \frac{40}{3} \approx 13.33 \).

Thus, the five arithmetic means are approximately **6.67, 8.33, 10, 11.67, and 13.33**.

---

Do you want details or have any further questions?

### 5 Related Questions:
1. How do you derive the formula for the sum of an arithmetic series?
2. What are some real-life applications of arithmetic sequences?
3. How do you find the common difference in an arithmetic sequence with only a few terms given?
4. Can an arithmetic sequence have negative common differences?
5. How do you calculate the sum of a geometric sequence?

### Tip:
When dealing with arithmetic sequences, always check if the context hints towards a common increase or decrease pattern to identify the common difference.

Solve/answer the following (show your solution in each problem): 1. One day, Isabela Jean jogged ½ kilometer. Because it was fun, she decided to increase the jogging distance each day by a certain amount. If she jogged 6 kilometers on the 51st day, how much more the distance increased each day? 2. Logs are stacked in a pile. The top row has 15 logs and the bottom has 21 logs. How many logs are in the stack? 3. Insert five arithmetic means between 5 and 15.

Solve arithmetic sequence problems involving Isabela's jogging distance, a log pile, and inserting arithmetic means. Learn how to use key formulas such as a_n = a_1 + (n - 1) * d to find the common difference and sum of sequences. Suitable for Grades 8-10 students.

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