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Sagot :
Certainly! Let's solve the two parts of the question step by step.
### Part 1: Finding the 31st term of an Arithmetic Progression (AP)
#### Given:
- The 11th term (\(a_{11}\)) is 38
- The 16th term (\(a_{16}\)) is 73
#### Steps to find the 31st term:
1. Recall the formula for the \(n\)-th term of an AP:
[tex]\[ a_n = a + (n - 1)d \][/tex]
where \(a\) is the first term and \(d\) is the common difference.
2. Write the equations for the 11th and 16th terms:
[tex]\[ a + 10d = 38 \quad \text{(1)} \][/tex]
[tex]\[ a + 15d = 73 \quad \text{(2)} \][/tex]
3. Subtract equation (1) from equation (2) to find \(d\):
[tex]\[ (a + 15d) - (a + 10d) = 73 - 38 \][/tex]
[tex]\[ 5d = 35 \][/tex]
[tex]\[ d = 7 \][/tex]
4. Substitute \(d = 7\) back into equation (1):
[tex]\[ a + 10 \cdot 7 = 38 \][/tex]
[tex]\[ a + 70 = 38 \][/tex]
[tex]\[ a = 38 - 70 \][/tex]
[tex]\[ a = -32 \][/tex]
5. Find the 31st term (\(a_{31}\)):
[tex]\[ a_{31} = a + 30d \][/tex]
[tex]\[ a_{31} = -32 + 30 \cdot 7 \][/tex]
[tex]\[ a_{31} = -32 + 210 \][/tex]
[tex]\[ a_{31} = 178 \][/tex]
Thus, the 31st term of the AP is \(178\).
### Part 2: Finding the term of the AP \(3, 15, 27, 39, \ldots\) which will be more than its 54th term
#### Given:
- The first term (\(a_2\)) is 3
- The common difference (\(d_2\)) is obtained from:
[tex]\[ d_2 = 15 - 3 = 12 \][/tex]
#### Steps to find the required term:
1. Calculate the 54th term (\(a_{54}\)):
[tex]\[ a_{54} = a_2 + 53d_2 \][/tex]
[tex]\[ a_{54} = 3 + 53 \cdot 12 \][/tex]
[tex]\[ a_{54} = 3 + 636 \][/tex]
[tex]\[ a_{54} = 639 \][/tex]
2. Find the smallest term number \(n\) such that \(a_n\) is greater than \(639\):
[tex]\[ a_n > 639 \][/tex]
[tex]\[ a_2 + (n - 1)d_2 > 639 \][/tex]
[tex]\[ 3 + (n - 1) \cdot 12 > 639 \][/tex]
[tex]\[ (n - 1) \cdot 12 > 639 - 3 \][/tex]
[tex]\[ (n - 1) \cdot 12 > 636 \][/tex]
[tex]\[ n - 1 > 636 / 12 \][/tex]
[tex]\[ n - 1 > 53 \][/tex]
[tex]\[ n > 53 + 1 \][/tex]
[tex]\[ n > 54 \][/tex]
Therefore, the 55th term of the AP \(3, 15, 27, 39, \ldots\) will be the first term to be more than its 54th term.
### Summary:
1. The 31st term of the AP whose 11th term is 38 and 16th term is 73 is \(178\).
2. The 55th term of the AP [tex]\(3, 15, 27, 39, \ldots\)[/tex] will be more than its 54th term.
### Part 1: Finding the 31st term of an Arithmetic Progression (AP)
#### Given:
- The 11th term (\(a_{11}\)) is 38
- The 16th term (\(a_{16}\)) is 73
#### Steps to find the 31st term:
1. Recall the formula for the \(n\)-th term of an AP:
[tex]\[ a_n = a + (n - 1)d \][/tex]
where \(a\) is the first term and \(d\) is the common difference.
2. Write the equations for the 11th and 16th terms:
[tex]\[ a + 10d = 38 \quad \text{(1)} \][/tex]
[tex]\[ a + 15d = 73 \quad \text{(2)} \][/tex]
3. Subtract equation (1) from equation (2) to find \(d\):
[tex]\[ (a + 15d) - (a + 10d) = 73 - 38 \][/tex]
[tex]\[ 5d = 35 \][/tex]
[tex]\[ d = 7 \][/tex]
4. Substitute \(d = 7\) back into equation (1):
[tex]\[ a + 10 \cdot 7 = 38 \][/tex]
[tex]\[ a + 70 = 38 \][/tex]
[tex]\[ a = 38 - 70 \][/tex]
[tex]\[ a = -32 \][/tex]
5. Find the 31st term (\(a_{31}\)):
[tex]\[ a_{31} = a + 30d \][/tex]
[tex]\[ a_{31} = -32 + 30 \cdot 7 \][/tex]
[tex]\[ a_{31} = -32 + 210 \][/tex]
[tex]\[ a_{31} = 178 \][/tex]
Thus, the 31st term of the AP is \(178\).
### Part 2: Finding the term of the AP \(3, 15, 27, 39, \ldots\) which will be more than its 54th term
#### Given:
- The first term (\(a_2\)) is 3
- The common difference (\(d_2\)) is obtained from:
[tex]\[ d_2 = 15 - 3 = 12 \][/tex]
#### Steps to find the required term:
1. Calculate the 54th term (\(a_{54}\)):
[tex]\[ a_{54} = a_2 + 53d_2 \][/tex]
[tex]\[ a_{54} = 3 + 53 \cdot 12 \][/tex]
[tex]\[ a_{54} = 3 + 636 \][/tex]
[tex]\[ a_{54} = 639 \][/tex]
2. Find the smallest term number \(n\) such that \(a_n\) is greater than \(639\):
[tex]\[ a_n > 639 \][/tex]
[tex]\[ a_2 + (n - 1)d_2 > 639 \][/tex]
[tex]\[ 3 + (n - 1) \cdot 12 > 639 \][/tex]
[tex]\[ (n - 1) \cdot 12 > 639 - 3 \][/tex]
[tex]\[ (n - 1) \cdot 12 > 636 \][/tex]
[tex]\[ n - 1 > 636 / 12 \][/tex]
[tex]\[ n - 1 > 53 \][/tex]
[tex]\[ n > 53 + 1 \][/tex]
[tex]\[ n > 54 \][/tex]
Therefore, the 55th term of the AP \(3, 15, 27, 39, \ldots\) will be the first term to be more than its 54th term.
### Summary:
1. The 31st term of the AP whose 11th term is 38 and 16th term is 73 is \(178\).
2. The 55th term of the AP [tex]\(3, 15, 27, 39, \ldots\)[/tex] will be more than its 54th term.
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