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Sagot :
Let's break down each of the parts of the question step by step.
### Part b: Determine the common difference and complete the table
The given sequence is: 4, 1, -2, -5, -8, -11, -14, -17, -20, -23
#### Finding the common difference:
- To find the common difference ([tex]\( d \)[/tex]), we subtract the first term from the second term:
[tex]\[ d = \text{{second term}} - \text{{first term}} = 1 - 4 = -3 \][/tex]
#### Completing the table:
We already have the terms of the sequence from the question. So, the completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Term} (n) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Term} (a_n) & 4 & 1 & -2 & -5 & -8 & -11 & -14 & -17 & -20 & -23 \\ \hline \end{array} \][/tex]
So, the common difference [tex]\( d = -3 \)[/tex].
### Part c: Create an arithmetic sequence where [tex]\( a_5 \)[/tex] is 10
We need to find a new sequence where the 5th term ([tex]\( a_5 \)[/tex]) is 10. Given that the common difference ([tex]\( d \)[/tex]) for our sequences is -3.
#### Equation for the [tex]\( n \)[/tex]-th term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Given:
- [tex]\( a_5 = 10 \)[/tex]
- Common difference [tex]\( d = -3 \)[/tex]
- [tex]\( n = 5 \)[/tex]
To find the first term ([tex]\( a_1 \)[/tex]), we use the formula:
[tex]\[ a_5 = a_1 + 4d \quad (since \ n-1 = 5-1 = 4) \][/tex]
Substitute [tex]\( a_5 = 10 \)[/tex] and [tex]\( d = -3 \)[/tex] into the equation:
[tex]\[ 10 = a_1 + 4(-3) \][/tex]
[tex]\[ 10 = a_1 - 12 \][/tex]
[tex]\[ a_1 = 10 + 12 \][/tex]
[tex]\[ a_1 = 22 \][/tex]
Now we can generate the sequence using [tex]\( a_1 = 22 \)[/tex] and [tex]\( d = -3 \)[/tex]:
[tex]\[ \begin{aligned} a_1 &= 22 \\ a_2 &= 22 - 3 = 19 \\ a_3 &= 19 - 3 = 16 \\ a_4 &= 16 - 3 = 13 \\ a_5 &= 13 - 3 = 10 \\ a_6 &= 10 - 3 = 7 \\ a_7 &= 7 - 3 = 4 \\ a_8 &= 4 - 3 = 1 \\ a_9 &= 1 - 3 = -2 \\ a_{10} &= -2 - 3 = -5 \\ \end{aligned} \][/tex]
So, the new sequence is: 22, 19, 16, 13, 10, 7, 4, 1, -2, -5
### Part d: Write an explicit equation model for the sequence in part a
The nth term for an arithmetic sequence is given by:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
For the sequence in part a:
- First term ([tex]\( a_1 \)[/tex]) is 4
- Common difference ([tex]\( d \)[/tex]) is -3
Therefore, the explicit equation is:
[tex]\[ a_n = 4 - 3(n-1) \][/tex]
In summary:
- Common difference ([tex]\( d \)[/tex]) is -3
- Completed table for the given terms:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Term} (n) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Term} (a_n) & 4 & 1 & -2 & -5 & -8 & -11 & -14 & -17 & -20 & -23 \\ \hline \end{array} \][/tex]
- New sequence with [tex]\( a_5 = 10 \)[/tex]: 22, 19, 16, 13, 10, 7, 4, 1, -2, -5
- Equation model for the initial sequence: [tex]\( a_n = 4 - 3(n-1) \)[/tex]
### Part b: Determine the common difference and complete the table
The given sequence is: 4, 1, -2, -5, -8, -11, -14, -17, -20, -23
#### Finding the common difference:
- To find the common difference ([tex]\( d \)[/tex]), we subtract the first term from the second term:
[tex]\[ d = \text{{second term}} - \text{{first term}} = 1 - 4 = -3 \][/tex]
#### Completing the table:
We already have the terms of the sequence from the question. So, the completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Term} (n) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Term} (a_n) & 4 & 1 & -2 & -5 & -8 & -11 & -14 & -17 & -20 & -23 \\ \hline \end{array} \][/tex]
So, the common difference [tex]\( d = -3 \)[/tex].
### Part c: Create an arithmetic sequence where [tex]\( a_5 \)[/tex] is 10
We need to find a new sequence where the 5th term ([tex]\( a_5 \)[/tex]) is 10. Given that the common difference ([tex]\( d \)[/tex]) for our sequences is -3.
#### Equation for the [tex]\( n \)[/tex]-th term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Given:
- [tex]\( a_5 = 10 \)[/tex]
- Common difference [tex]\( d = -3 \)[/tex]
- [tex]\( n = 5 \)[/tex]
To find the first term ([tex]\( a_1 \)[/tex]), we use the formula:
[tex]\[ a_5 = a_1 + 4d \quad (since \ n-1 = 5-1 = 4) \][/tex]
Substitute [tex]\( a_5 = 10 \)[/tex] and [tex]\( d = -3 \)[/tex] into the equation:
[tex]\[ 10 = a_1 + 4(-3) \][/tex]
[tex]\[ 10 = a_1 - 12 \][/tex]
[tex]\[ a_1 = 10 + 12 \][/tex]
[tex]\[ a_1 = 22 \][/tex]
Now we can generate the sequence using [tex]\( a_1 = 22 \)[/tex] and [tex]\( d = -3 \)[/tex]:
[tex]\[ \begin{aligned} a_1 &= 22 \\ a_2 &= 22 - 3 = 19 \\ a_3 &= 19 - 3 = 16 \\ a_4 &= 16 - 3 = 13 \\ a_5 &= 13 - 3 = 10 \\ a_6 &= 10 - 3 = 7 \\ a_7 &= 7 - 3 = 4 \\ a_8 &= 4 - 3 = 1 \\ a_9 &= 1 - 3 = -2 \\ a_{10} &= -2 - 3 = -5 \\ \end{aligned} \][/tex]
So, the new sequence is: 22, 19, 16, 13, 10, 7, 4, 1, -2, -5
### Part d: Write an explicit equation model for the sequence in part a
The nth term for an arithmetic sequence is given by:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
For the sequence in part a:
- First term ([tex]\( a_1 \)[/tex]) is 4
- Common difference ([tex]\( d \)[/tex]) is -3
Therefore, the explicit equation is:
[tex]\[ a_n = 4 - 3(n-1) \][/tex]
In summary:
- Common difference ([tex]\( d \)[/tex]) is -3
- Completed table for the given terms:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Term} (n) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Term} (a_n) & 4 & 1 & -2 & -5 & -8 & -11 & -14 & -17 & -20 & -23 \\ \hline \end{array} \][/tex]
- New sequence with [tex]\( a_5 = 10 \)[/tex]: 22, 19, 16, 13, 10, 7, 4, 1, -2, -5
- Equation model for the initial sequence: [tex]\( a_n = 4 - 3(n-1) \)[/tex]
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