Answer :
* Finds the value of $x$ in the AP $x+3, 2x+1, x-7$ by setting the common differences equal: $x = -3$.
* Determines $a, b, c$ in the AP $a, 10, b, c, 31$ by finding the common difference: $a = 3, b = 17, c = 24$.
* Calculates the $21^{st}$ term of the AP $-4 \frac{1}{2}, -3, -1 \frac{1}{2}, \ldots$ using the formula $a_n = a + (n-1)d$: $25.5$.
* Finds the $7^{th}$ term from the end of the AP $7, 10, 13, \ldots, 184$ by determining the number of terms and working backwards: $166$.
* Identifies the middle term of the AP $213, 205, 197, \ldots, 37$ by finding the number of terms and selecting the middle one: $125$.
### Explanation
1. Problem Overview
We are given five separate arithmetic progression (AP) problems to solve. We will address each one individually, showing all steps clearly.
2. Finding x
1. If the numbers $x+3, 2x+1$ and $x-7$ are in AP, then the common difference between consecutive terms is the same. Thus, $(2x+1) - (x+3) = (x-7) - (2x+1)$. Simplifying this equation, we get $2x + 1 - x - 3 = x - 7 - 2x - 1$, which simplifies to $x - 2 = -x - 8$. Adding $x$ to both sides gives $2x - 2 = -8$. Adding 2 to both sides gives $2x = -6$. Dividing by 2 gives $x = -3$.
3. Finding a, b, and c
2. If the numbers $a, 10, b, c, 31$ are in AP, then the common difference is constant. Let the common difference be $d$. Then, $10 = a + d$, $b = 10 + d$, $c = b + d$, and $31 = c + d$. Also, $31 = a + 4d$. From $10 = a + d$, we have $a = 10 - d$. Substituting this into $31 = a + 4d$, we get $31 = (10 - d) + 4d$, which simplifies to $31 = 10 + 3d$. Subtracting 10 from both sides gives $21 = 3d$. Dividing by 3 gives $d = 7$. Thus, $a = 10 - 7 = 3$, $b = 10 + 7 = 17$, and $c = 17 + 7 = 24$.
4. Finding the 21st term
3. For the AP $-4
\frac{1}{2}, -3, -1
\frac{1}{2},
\ldots$, the first term is $a = -4
\frac{1}{2} = -4.5$ and the common difference is $d = -3 - (-4.5) = 1.5$. The $21^{st}$ term is given by $a_{21} = a + (21-1)d = -4.5 + 20(1.5) = -4.5 + 30 = 25.5$.
5. Finding the 7th term from the end
4. For the AP $7, 10, 13,
\ldots, 184$, the first term is $a = 7$ and the common difference is $d = 10 - 7 = 3$. To find the number of terms, we use the formula $l = a + (n-1)d$, where $l$ is the last term. Thus, $184 = 7 + (n-1)3$. Subtracting 7 from both sides gives $177 = (n-1)3$. Dividing by 3 gives $59 = n-1$, so $n = 60$. The $7^{th}$ term from the end is the $(60-7+1)^{th} = 54^{th}$ term from the beginning. Thus, the $54^{th}$ term is $a_{54} = a + (54-1)d = 7 + 53(3) = 7 + 159 = 166$.
6. Finding the middle term
5. For the AP $213, 205, 197,
\ldots, 37$, the first term is $a = 213$ and the common difference is $d = 205 - 213 = -8$. To find the number of terms, we use the formula $l = a + (n-1)d$, where $l$ is the last term. Thus, $37 = 213 + (n-1)(-8)$. Subtracting 213 from both sides gives $-176 = (n-1)(-8)$. Dividing by -8 gives $22 = n-1$, so $n = 23$. Since $n$ is odd, there is one middle term, which is the $(\frac{23+1}{2})^{th} = 12^{th}$ term. The $12^{th}$ term is $a_{12} = a + (12-1)d = 213 + 11(-8) = 213 - 88 = 125$.
7. Final Answers
Therefore, the solutions are:
1. $x = -3$
2. $a = 3, b = 17, c = 24$
3. $21^{st}$ term is $25.5$
4. $7^{th}$ term from the end is $166$
5. The middle term is $125$
### 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 a stadium with a consistent row increase. Understanding AP helps in making informed decisions in finance, planning, and resource allocation where quantities increase or decrease linearly.
* Determines $a, b, c$ in the AP $a, 10, b, c, 31$ by finding the common difference: $a = 3, b = 17, c = 24$.
* Calculates the $21^{st}$ term of the AP $-4 \frac{1}{2}, -3, -1 \frac{1}{2}, \ldots$ using the formula $a_n = a + (n-1)d$: $25.5$.
* Finds the $7^{th}$ term from the end of the AP $7, 10, 13, \ldots, 184$ by determining the number of terms and working backwards: $166$.
* Identifies the middle term of the AP $213, 205, 197, \ldots, 37$ by finding the number of terms and selecting the middle one: $125$.
### Explanation
1. Problem Overview
We are given five separate arithmetic progression (AP) problems to solve. We will address each one individually, showing all steps clearly.
2. Finding x
1. If the numbers $x+3, 2x+1$ and $x-7$ are in AP, then the common difference between consecutive terms is the same. Thus, $(2x+1) - (x+3) = (x-7) - (2x+1)$. Simplifying this equation, we get $2x + 1 - x - 3 = x - 7 - 2x - 1$, which simplifies to $x - 2 = -x - 8$. Adding $x$ to both sides gives $2x - 2 = -8$. Adding 2 to both sides gives $2x = -6$. Dividing by 2 gives $x = -3$.
3. Finding a, b, and c
2. If the numbers $a, 10, b, c, 31$ are in AP, then the common difference is constant. Let the common difference be $d$. Then, $10 = a + d$, $b = 10 + d$, $c = b + d$, and $31 = c + d$. Also, $31 = a + 4d$. From $10 = a + d$, we have $a = 10 - d$. Substituting this into $31 = a + 4d$, we get $31 = (10 - d) + 4d$, which simplifies to $31 = 10 + 3d$. Subtracting 10 from both sides gives $21 = 3d$. Dividing by 3 gives $d = 7$. Thus, $a = 10 - 7 = 3$, $b = 10 + 7 = 17$, and $c = 17 + 7 = 24$.
4. Finding the 21st term
3. For the AP $-4
\frac{1}{2}, -3, -1
\frac{1}{2},
\ldots$, the first term is $a = -4
\frac{1}{2} = -4.5$ and the common difference is $d = -3 - (-4.5) = 1.5$. The $21^{st}$ term is given by $a_{21} = a + (21-1)d = -4.5 + 20(1.5) = -4.5 + 30 = 25.5$.
5. Finding the 7th term from the end
4. For the AP $7, 10, 13,
\ldots, 184$, the first term is $a = 7$ and the common difference is $d = 10 - 7 = 3$. To find the number of terms, we use the formula $l = a + (n-1)d$, where $l$ is the last term. Thus, $184 = 7 + (n-1)3$. Subtracting 7 from both sides gives $177 = (n-1)3$. Dividing by 3 gives $59 = n-1$, so $n = 60$. The $7^{th}$ term from the end is the $(60-7+1)^{th} = 54^{th}$ term from the beginning. Thus, the $54^{th}$ term is $a_{54} = a + (54-1)d = 7 + 53(3) = 7 + 159 = 166$.
6. Finding the middle term
5. For the AP $213, 205, 197,
\ldots, 37$, the first term is $a = 213$ and the common difference is $d = 205 - 213 = -8$. To find the number of terms, we use the formula $l = a + (n-1)d$, where $l$ is the last term. Thus, $37 = 213 + (n-1)(-8)$. Subtracting 213 from both sides gives $-176 = (n-1)(-8)$. Dividing by -8 gives $22 = n-1$, so $n = 23$. Since $n$ is odd, there is one middle term, which is the $(\frac{23+1}{2})^{th} = 12^{th}$ term. The $12^{th}$ term is $a_{12} = a + (12-1)d = 213 + 11(-8) = 213 - 88 = 125$.
7. Final Answers
Therefore, the solutions are:
1. $x = -3$
2. $a = 3, b = 17, c = 24$
3. $21^{st}$ term is $25.5$
4. $7^{th}$ term from the end is $166$
5. The middle term is $125$
### 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 a stadium with a consistent row increase. Understanding AP helps in making informed decisions in finance, planning, and resource allocation where quantities increase or decrease linearly.
