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Answer :
- For the first problem, form two equations based on the sums of the first 7 terms and the next 7 terms.
- Solve the system of equations to find the first term $a$ and the common difference $d$.
- For the second problem, form two equations based on the given relationships between the 5th, 9th, 25th, and 8th terms.
- Solve the system of equations to find the first term $a$ and the common difference $d$.
- The AP for both problems is: $\boxed{3, 5, 7, 9, ...}$
### Explanation
1. Understanding the Problem
We are given two problems, both asking us to find an Arithmetic Progression (AP). An AP is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by 'd'. The nth term of an AP is given by $a_n = a + (n-1)d$, where 'a' is the first term. The sum of the first n terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
2. Setting up equations for Problem 1
**Problem 1:** The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Let 'a' be the first term and 'd' be the common difference. The sum of the first 7 terms is 63: $S_7 = \frac{7}{2}[2a + 6d] = 63$. Simplifying this equation, we get $2a + 6d = 18$ or $a + 3d = 9$. The sum of the next 7 terms is 161. This means the sum of the first 14 terms minus the sum of the first 7 terms is 161. So, $S_{14} - S_7 = 161$. We know $S_7 = 63$, so $S_{14} = 161 + 63 = 224$. Therefore, $S_{14} = \frac{14}{2}[2a + 13d] = 224$. Simplifying this equation, we get $7[2a + 13d] = 224$ or $2a + 13d = 32$. Now we have a system of equations: $a + 3d = 9$ and $2a + 13d = 32$.
3. Solving Problem 1
Multiplying the first equation by 2, we get $2a + 6d = 18$. Subtracting this from the second equation ($2a + 13d = 32$), we get $7d = 14$, so $d = 2$. Substituting $d = 2$ into $a + 3d = 9$, we get $a + 6 = 9$, so $a = 3$. Therefore, the AP for the first problem is 3, 5, 7, 9, ...
4. Setting up equations for Problem 2
**Problem 2:** The sum of the $5^{\text{th}}$ and the $9^{\text{th}}$ terms of an AP is 30. If its $25^{\text{th}}$ term is three times its $8^{\text{th}}$ term, find the AP. Let 'a' be the first term and 'd' be the common difference. The sum of the 5th and 9th terms is 30: $a_5 + a_9 = 30$. This means $(a + 4d) + (a + 8d) = 30$, which simplifies to $2a + 12d = 30$ or $a + 6d = 15$. The 25th term is three times the 8th term: $a_{25} = 3a_8$. This means $a + 24d = 3(a + 7d)$, which simplifies to $a + 24d = 3a + 21d$ or $2a - 3d = 0$. Now we have a system of equations: $a + 6d = 15$ and $2a - 3d = 0$.
5. Solving Problem 2
Multiplying the first equation by 2, we get $2a + 12d = 30$. Subtracting the second equation ($2a - 3d = 0$) from this, we get $15d = 30$, so $d = 2$. Substituting $d = 2$ into $a + 6d = 15$, we get $a + 12 = 15$, so $a = 3$. Therefore, the AP for the second problem is 3, 5, 7, 9, ...
6. Final Answer
Therefore, the Arithmetic Progression for both problems is the same: 3, 5, 7, 9, ...
### Examples
Arithmetic progressions are useful in various real-life scenarios, such as calculating simple interest, predicting salary increases, or determining the number of seats in rows of a theater. Understanding and solving AP problems helps in making predictions and planning in situations where quantities increase at a constant rate. For instance, if you deposit a fixed amount of money into a savings account each month, the total amount saved over time forms an arithmetic progression.
- Solve the system of equations to find the first term $a$ and the common difference $d$.
- For the second problem, form two equations based on the given relationships between the 5th, 9th, 25th, and 8th terms.
- Solve the system of equations to find the first term $a$ and the common difference $d$.
- The AP for both problems is: $\boxed{3, 5, 7, 9, ...}$
### Explanation
1. Understanding the Problem
We are given two problems, both asking us to find an Arithmetic Progression (AP). An AP is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by 'd'. The nth term of an AP is given by $a_n = a + (n-1)d$, where 'a' is the first term. The sum of the first n terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
2. Setting up equations for Problem 1
**Problem 1:** The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Let 'a' be the first term and 'd' be the common difference. The sum of the first 7 terms is 63: $S_7 = \frac{7}{2}[2a + 6d] = 63$. Simplifying this equation, we get $2a + 6d = 18$ or $a + 3d = 9$. The sum of the next 7 terms is 161. This means the sum of the first 14 terms minus the sum of the first 7 terms is 161. So, $S_{14} - S_7 = 161$. We know $S_7 = 63$, so $S_{14} = 161 + 63 = 224$. Therefore, $S_{14} = \frac{14}{2}[2a + 13d] = 224$. Simplifying this equation, we get $7[2a + 13d] = 224$ or $2a + 13d = 32$. Now we have a system of equations: $a + 3d = 9$ and $2a + 13d = 32$.
3. Solving Problem 1
Multiplying the first equation by 2, we get $2a + 6d = 18$. Subtracting this from the second equation ($2a + 13d = 32$), we get $7d = 14$, so $d = 2$. Substituting $d = 2$ into $a + 3d = 9$, we get $a + 6 = 9$, so $a = 3$. Therefore, the AP for the first problem is 3, 5, 7, 9, ...
4. Setting up equations for Problem 2
**Problem 2:** The sum of the $5^{\text{th}}$ and the $9^{\text{th}}$ terms of an AP is 30. If its $25^{\text{th}}$ term is three times its $8^{\text{th}}$ term, find the AP. Let 'a' be the first term and 'd' be the common difference. The sum of the 5th and 9th terms is 30: $a_5 + a_9 = 30$. This means $(a + 4d) + (a + 8d) = 30$, which simplifies to $2a + 12d = 30$ or $a + 6d = 15$. The 25th term is three times the 8th term: $a_{25} = 3a_8$. This means $a + 24d = 3(a + 7d)$, which simplifies to $a + 24d = 3a + 21d$ or $2a - 3d = 0$. Now we have a system of equations: $a + 6d = 15$ and $2a - 3d = 0$.
5. Solving Problem 2
Multiplying the first equation by 2, we get $2a + 12d = 30$. Subtracting the second equation ($2a - 3d = 0$) from this, we get $15d = 30$, so $d = 2$. Substituting $d = 2$ into $a + 6d = 15$, we get $a + 12 = 15$, so $a = 3$. Therefore, the AP for the second problem is 3, 5, 7, 9, ...
6. Final Answer
Therefore, the Arithmetic Progression for both problems is the same: 3, 5, 7, 9, ...
### Examples
Arithmetic progressions are useful in various real-life scenarios, such as calculating simple interest, predicting salary increases, or determining the number of seats in rows of a theater. Understanding and solving AP problems helps in making predictions and planning in situations where quantities increase at a constant rate. For instance, if you deposit a fixed amount of money into a savings account each month, the total amount saved over time forms an arithmetic progression.
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