Answer :
- For question 15, the common difference $d$ is found by using the formula for the nth term of an AP and the given condition $a_{17} - a_{10} = 7$, resulting in $d = 1$.
- For question 16, we use the formula for the nth term of an AP and the given condition $5a_5 = 8a_8$ to find a relationship between $a_1$ and $d$, which is $a_1 = -12d$.
- Substituting this relationship into the formula for the 13th term, $a_{13} = a_1 + 12d$, we find that $a_{13} = 0$.
- Therefore, the common difference for question 15 is $\boxed{1}$, and the 13th term for question 16 is $\boxed{0}$.
### Explanation
1. Problem Analysis
We are given two problems related to arithmetic progressions (AP). In the first problem, we need to find the common difference of an AP where the 17th term exceeds the 10th term by 7. In the second problem, we need to show that the 13th term of an AP is zero, given that five times the fifth term is equal to eight times the eighth term.
2. Question 15: Setting up the equation
For question 15, let $a_{17}$ be the 17th term and $a_{10}$ be the 10th term. We are given that $a_{17} - a_{10} = 7$. The general formula for the nth term of an AP is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. Therefore, $a_{17} = a_1 + 16d$ and $a_{10} = a_1 + 9d$. Substituting these into the given equation, we get $(a_1 + 16d) - (a_1 + 9d) = 7$.
3. Question 15: Solving for d
Simplifying the equation $(a_1 + 16d) - (a_1 + 9d) = 7$, we have $7d = 7$. Dividing both sides by 7, we get $d = 1$. Therefore, the common difference is 1.
4. Question 16: Setting up the equation
For question 16, let $a_5$ be the 5th term and $a_8$ be the 8th term. We are given that $5a_5 = 8a_8$. We need to show that $a_{13} = 0$. Using the general formula for the nth term, we have $a_5 = a_1 + 4d$ and $a_8 = a_1 + 7d$. Substituting these into the given equation, we get $5(a_1 + 4d) = 8(a_1 + 7d)$.
5. Question 16: Showing that a_13 = 0
Expanding the equation $5(a_1 + 4d) = 8(a_1 + 7d)$, we get $5a_1 + 20d = 8a_1 + 56d$. Rearranging the terms, we have $3a_1 = -36d$, which simplifies to $a_1 = -12d$. Now, we need to find $a_{13}$. Using the general formula, $a_{13} = a_1 + 12d$. Substituting $a_1 = -12d$ into this equation, we get $a_{13} = -12d + 12d = 0$. Therefore, the 13th term is zero.
6. Final Answers
In summary, for question 15, the common difference is 1. For question 16, we have shown that the 13th term is zero.
### Examples
Arithmetic progressions are useful in various real-life scenarios, such as calculating simple interest, predicting salary increases, and modeling the depreciation of assets. Understanding and solving problems related to APs can help in making informed financial decisions and predicting future trends.
- For question 16, we use the formula for the nth term of an AP and the given condition $5a_5 = 8a_8$ to find a relationship between $a_1$ and $d$, which is $a_1 = -12d$.
- Substituting this relationship into the formula for the 13th term, $a_{13} = a_1 + 12d$, we find that $a_{13} = 0$.
- Therefore, the common difference for question 15 is $\boxed{1}$, and the 13th term for question 16 is $\boxed{0}$.
### Explanation
1. Problem Analysis
We are given two problems related to arithmetic progressions (AP). In the first problem, we need to find the common difference of an AP where the 17th term exceeds the 10th term by 7. In the second problem, we need to show that the 13th term of an AP is zero, given that five times the fifth term is equal to eight times the eighth term.
2. Question 15: Setting up the equation
For question 15, let $a_{17}$ be the 17th term and $a_{10}$ be the 10th term. We are given that $a_{17} - a_{10} = 7$. The general formula for the nth term of an AP is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. Therefore, $a_{17} = a_1 + 16d$ and $a_{10} = a_1 + 9d$. Substituting these into the given equation, we get $(a_1 + 16d) - (a_1 + 9d) = 7$.
3. Question 15: Solving for d
Simplifying the equation $(a_1 + 16d) - (a_1 + 9d) = 7$, we have $7d = 7$. Dividing both sides by 7, we get $d = 1$. Therefore, the common difference is 1.
4. Question 16: Setting up the equation
For question 16, let $a_5$ be the 5th term and $a_8$ be the 8th term. We are given that $5a_5 = 8a_8$. We need to show that $a_{13} = 0$. Using the general formula for the nth term, we have $a_5 = a_1 + 4d$ and $a_8 = a_1 + 7d$. Substituting these into the given equation, we get $5(a_1 + 4d) = 8(a_1 + 7d)$.
5. Question 16: Showing that a_13 = 0
Expanding the equation $5(a_1 + 4d) = 8(a_1 + 7d)$, we get $5a_1 + 20d = 8a_1 + 56d$. Rearranging the terms, we have $3a_1 = -36d$, which simplifies to $a_1 = -12d$. Now, we need to find $a_{13}$. Using the general formula, $a_{13} = a_1 + 12d$. Substituting $a_1 = -12d$ into this equation, we get $a_{13} = -12d + 12d = 0$. Therefore, the 13th term is zero.
6. Final Answers
In summary, for question 15, the common difference is 1. For question 16, we have shown that the 13th term is zero.
### Examples
Arithmetic progressions are useful in various real-life scenarios, such as calculating simple interest, predicting salary increases, and modeling the depreciation of assets. Understanding and solving problems related to APs can help in making informed financial decisions and predicting future trends.
