Answer :
Sure! Let's go through each problem one by one with detailed explanations.
Problem 26:
Find the first term of an arithmetic sequence where the fourth term is 8 and the common difference is 2.
1. The formula for the nth term of an arithmetic sequence is:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
where [tex]\(a_n\)[/tex] is the nth term, [tex]\(a_1\)[/tex] is the first term, [tex]\(n\)[/tex] is the term number, and [tex]\(d\)[/tex] is the common difference.
2. We know [tex]\(a_4 = 8\)[/tex] and [tex]\(d = 2\)[/tex]. So, we set up the equation:
[tex]\[ 8 = a_1 + (4-1) \cdot 2 \][/tex]
[tex]\[ 8 = a_1 + 6 \][/tex]
3. Solving for [tex]\(a_1\)[/tex]:
[tex]\[ a_1 = 8 - 6 \][/tex]
[tex]\[ a_1 = 2 \][/tex]
The first term of the sequence is 2.
Problem 27:
Find the first term of an arithmetic sequence where the 12th term is 22 and the common difference is -5.
1. Using the nth term formula again:
[tex]\[ a_{12} = a_1 + (12-1) \cdot (-5) \][/tex]
[tex]\[ 22 = a_1 - 55 \][/tex]
2. Solving for [tex]\(a_1\)[/tex]:
[tex]\[ a_1 = 22 + 55 \][/tex]
[tex]\[ a_1 = 77 \][/tex]
The first term of the sequence is 77.
Problem 28:
Find which term in the sequence 28, 20, 12, 4, ... is -36.
1. First, identify the common difference:
[tex]\[ 20 - 28 = -8 \][/tex]
2. Use the nth term formula to find the desired term:
[tex]\[ a_n = a_1 + (n-1) \cdot (-8) = -36 \][/tex]
[tex]\[ -36 = 28 + (n-1) \cdot (-8) \][/tex]
[tex]\[ -36 = 28 - 8n + 8 \][/tex]
[tex]\[ -36 = 36 - 8n \][/tex]
3. Solving for [tex]\(n\)[/tex]:
[tex]\[ -8n = -36 - 36 \][/tex]
[tex]\[ -8n = -72 \][/tex]
[tex]\[ n = 9 \][/tex]
The -36 is the 9th term in the sequence.
Problem 29:
Calculate the odometer reading after 25 days if the initial reading is 24,521 miles and Jamal drives 45 miles every day.
1. Calculate the total miles driven in 25 days:
[tex]\[ 45 \cdot 25 = 1125 \][/tex]
2. Add this to the initial odometer reading:
[tex]\[ 24,521 + 1125 = 25,646 \][/tex]
The odometer reading after 25 days will be 25,646 miles.
Problem 30:
Determine the function representing the sequence, the weight of each yearbook, and the number of yearbooks in the box.
a. Function:
The sequence is 281, 270, 259, 248, ..., with a common difference of -11. The function can be expressed as:
[tex]\[ f(n) = 281 - 11(n-1) \][/tex]
b. Weight of each yearbook:
Since the common difference is -11, the weight of each yearbook is 11 ounces.
c. Number of yearbooks:
Given the box weighs 292 ounces full and at least 17 ounces empty:
[tex]\[ 292 - 17 = 275 \][/tex]
Divide the weight difference by the weight of each yearbook to find the number of yearbooks:
[tex]\[ \frac{275}{11} = 25 \][/tex]
There are 25 yearbooks in the box.
Problem 26:
Find the first term of an arithmetic sequence where the fourth term is 8 and the common difference is 2.
1. The formula for the nth term of an arithmetic sequence is:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
where [tex]\(a_n\)[/tex] is the nth term, [tex]\(a_1\)[/tex] is the first term, [tex]\(n\)[/tex] is the term number, and [tex]\(d\)[/tex] is the common difference.
2. We know [tex]\(a_4 = 8\)[/tex] and [tex]\(d = 2\)[/tex]. So, we set up the equation:
[tex]\[ 8 = a_1 + (4-1) \cdot 2 \][/tex]
[tex]\[ 8 = a_1 + 6 \][/tex]
3. Solving for [tex]\(a_1\)[/tex]:
[tex]\[ a_1 = 8 - 6 \][/tex]
[tex]\[ a_1 = 2 \][/tex]
The first term of the sequence is 2.
Problem 27:
Find the first term of an arithmetic sequence where the 12th term is 22 and the common difference is -5.
1. Using the nth term formula again:
[tex]\[ a_{12} = a_1 + (12-1) \cdot (-5) \][/tex]
[tex]\[ 22 = a_1 - 55 \][/tex]
2. Solving for [tex]\(a_1\)[/tex]:
[tex]\[ a_1 = 22 + 55 \][/tex]
[tex]\[ a_1 = 77 \][/tex]
The first term of the sequence is 77.
Problem 28:
Find which term in the sequence 28, 20, 12, 4, ... is -36.
1. First, identify the common difference:
[tex]\[ 20 - 28 = -8 \][/tex]
2. Use the nth term formula to find the desired term:
[tex]\[ a_n = a_1 + (n-1) \cdot (-8) = -36 \][/tex]
[tex]\[ -36 = 28 + (n-1) \cdot (-8) \][/tex]
[tex]\[ -36 = 28 - 8n + 8 \][/tex]
[tex]\[ -36 = 36 - 8n \][/tex]
3. Solving for [tex]\(n\)[/tex]:
[tex]\[ -8n = -36 - 36 \][/tex]
[tex]\[ -8n = -72 \][/tex]
[tex]\[ n = 9 \][/tex]
The -36 is the 9th term in the sequence.
Problem 29:
Calculate the odometer reading after 25 days if the initial reading is 24,521 miles and Jamal drives 45 miles every day.
1. Calculate the total miles driven in 25 days:
[tex]\[ 45 \cdot 25 = 1125 \][/tex]
2. Add this to the initial odometer reading:
[tex]\[ 24,521 + 1125 = 25,646 \][/tex]
The odometer reading after 25 days will be 25,646 miles.
Problem 30:
Determine the function representing the sequence, the weight of each yearbook, and the number of yearbooks in the box.
a. Function:
The sequence is 281, 270, 259, 248, ..., with a common difference of -11. The function can be expressed as:
[tex]\[ f(n) = 281 - 11(n-1) \][/tex]
b. Weight of each yearbook:
Since the common difference is -11, the weight of each yearbook is 11 ounces.
c. Number of yearbooks:
Given the box weighs 292 ounces full and at least 17 ounces empty:
[tex]\[ 292 - 17 = 275 \][/tex]
Divide the weight difference by the weight of each yearbook to find the number of yearbooks:
[tex]\[ \frac{275}{11} = 25 \][/tex]
There are 25 yearbooks in the box.
