Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
