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Sagot :
## Solution:
### 1. [tex]\(a_2 = 24\)[/tex] and [tex]\(a_5 = 3\)[/tex]. Find [tex]\(a_1\)[/tex] and [tex]\(d\)[/tex].
Using the [tex]\(n\)[/tex]th term formula for an arithmetic sequence:
[tex]\[a_n = a_1 + (n-1)d\][/tex]
For [tex]\(a_2\)[/tex]:
[tex]\[a_2 = a_1 + (2-1)d\][/tex]
[tex]\[24 = a_1 + d \quad \text{(Equation 1)}\][/tex]
For [tex]\(a_5\)[/tex]:
[tex]\[a_5 = a_1 + (5-1)d\][/tex]
[tex]\[3 = a_1 + 4d \quad \text{(Equation 2)}\][/tex]
Subtract Equation 1 from Equation 2:
[tex]\[ a_1 + 4d - (a_1 + d) = 3 - 24 \][/tex]
[tex]\[ 3d = -21 \][/tex]
[tex]\[ d = -7 \][/tex]
Substitute [tex]\(d = -7\)[/tex] back into Equation 1:
[tex]\[ 24 = a_1 + (-7) \][/tex]
[tex]\[ a_1 = 24 + 7 \][/tex]
[tex]\[ a_1 = 31 \][/tex]
So, [tex]\(a_1 = 31\)[/tex] and [tex]\(d = -7\)[/tex].
### 2. [tex]\(a_1 = 54\)[/tex] and [tex]\(a_9 = 60\)[/tex]. Find [tex]\(d\)[/tex].
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = 54 + (9-1)d\][/tex]
[tex]\[60 = 54 + 8d\][/tex]
Solve for [tex]\(d\)[/tex]:
[tex]\[ 60 - 54 = 8d \][/tex]
[tex]\[ 6 = 8d \][/tex]
[tex]\[ d = \frac{6}{8} \][/tex]
[tex]\[ d = \frac{3}{4} \][/tex]
Since [tex]\(a_1 = 54\)[/tex] is already given, there's no further calculation required for [tex]\(a_1\)[/tex].
So, [tex]\(d = \frac{3}{4}\)[/tex].
### 3. Which term of the arithmetic sequence is -18, given that [tex]\(a_1 = 7\)[/tex] and [tex]\(a_2 = 2\)[/tex].
First, find the common difference [tex]\(d\)[/tex]:
[tex]\[d = a_2 - a_1 = 2 - 7 = -5\][/tex]
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = 7 + (n-1)(-5)\][/tex]
Set [tex]\(a_n\)[/tex] to -18:
[tex]\[ -18 = 7 + (n-1)(-5) \][/tex]
[tex]\[ -18 = 7 - 5n + 5 \][/tex]
[tex]\[ -18 = 12 - 5n \][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[ -18 - 12 = -5n \][/tex]
[tex]\[ -30 = -5n \][/tex]
[tex]\[ n = \frac{30}{5} \][/tex]
[tex]\[ n = 6 \][/tex]
So, the term at which the value is -18 is the 6th term.
### 4. How many terms are in an arithmetic sequence whose first term is -3, common difference is 2, and the last term is 23.
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = -3 + (n-1)2 = 23\][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[ 23 = -3 + (n-1)2 \][/tex]
[tex]\[ 23 + 3 = 2(n-1) \][/tex]
[tex]\[ 26 = 2(n-1) \][/tex]
[tex]\[ 26 = 2n - 2 \][/tex]
[tex]\[ 26 + 2 = 2n \][/tex]
[tex]\[ 28 = 2n \][/tex]
[tex]\[ n = \frac{28}{2} \][/tex]
[tex]\[ n = 14 \][/tex]
So, there are 14 terms in the sequence.
## Summary of Answers:
1. [tex]\(a_1 = 31\)[/tex] and [tex]\(d = -7\)[/tex]
2. [tex]\(a_1 = 54\)[/tex]
3. The term at which the value is -18 is the 6th term.
4. There are 14 terms in the sequence.
### 1. [tex]\(a_2 = 24\)[/tex] and [tex]\(a_5 = 3\)[/tex]. Find [tex]\(a_1\)[/tex] and [tex]\(d\)[/tex].
Using the [tex]\(n\)[/tex]th term formula for an arithmetic sequence:
[tex]\[a_n = a_1 + (n-1)d\][/tex]
For [tex]\(a_2\)[/tex]:
[tex]\[a_2 = a_1 + (2-1)d\][/tex]
[tex]\[24 = a_1 + d \quad \text{(Equation 1)}\][/tex]
For [tex]\(a_5\)[/tex]:
[tex]\[a_5 = a_1 + (5-1)d\][/tex]
[tex]\[3 = a_1 + 4d \quad \text{(Equation 2)}\][/tex]
Subtract Equation 1 from Equation 2:
[tex]\[ a_1 + 4d - (a_1 + d) = 3 - 24 \][/tex]
[tex]\[ 3d = -21 \][/tex]
[tex]\[ d = -7 \][/tex]
Substitute [tex]\(d = -7\)[/tex] back into Equation 1:
[tex]\[ 24 = a_1 + (-7) \][/tex]
[tex]\[ a_1 = 24 + 7 \][/tex]
[tex]\[ a_1 = 31 \][/tex]
So, [tex]\(a_1 = 31\)[/tex] and [tex]\(d = -7\)[/tex].
### 2. [tex]\(a_1 = 54\)[/tex] and [tex]\(a_9 = 60\)[/tex]. Find [tex]\(d\)[/tex].
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = 54 + (9-1)d\][/tex]
[tex]\[60 = 54 + 8d\][/tex]
Solve for [tex]\(d\)[/tex]:
[tex]\[ 60 - 54 = 8d \][/tex]
[tex]\[ 6 = 8d \][/tex]
[tex]\[ d = \frac{6}{8} \][/tex]
[tex]\[ d = \frac{3}{4} \][/tex]
Since [tex]\(a_1 = 54\)[/tex] is already given, there's no further calculation required for [tex]\(a_1\)[/tex].
So, [tex]\(d = \frac{3}{4}\)[/tex].
### 3. Which term of the arithmetic sequence is -18, given that [tex]\(a_1 = 7\)[/tex] and [tex]\(a_2 = 2\)[/tex].
First, find the common difference [tex]\(d\)[/tex]:
[tex]\[d = a_2 - a_1 = 2 - 7 = -5\][/tex]
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = 7 + (n-1)(-5)\][/tex]
Set [tex]\(a_n\)[/tex] to -18:
[tex]\[ -18 = 7 + (n-1)(-5) \][/tex]
[tex]\[ -18 = 7 - 5n + 5 \][/tex]
[tex]\[ -18 = 12 - 5n \][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[ -18 - 12 = -5n \][/tex]
[tex]\[ -30 = -5n \][/tex]
[tex]\[ n = \frac{30}{5} \][/tex]
[tex]\[ n = 6 \][/tex]
So, the term at which the value is -18 is the 6th term.
### 4. How many terms are in an arithmetic sequence whose first term is -3, common difference is 2, and the last term is 23.
Using the [tex]\(n\)[/tex]th term formula:
[tex]\[a_n = -3 + (n-1)2 = 23\][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[ 23 = -3 + (n-1)2 \][/tex]
[tex]\[ 23 + 3 = 2(n-1) \][/tex]
[tex]\[ 26 = 2(n-1) \][/tex]
[tex]\[ 26 = 2n - 2 \][/tex]
[tex]\[ 26 + 2 = 2n \][/tex]
[tex]\[ 28 = 2n \][/tex]
[tex]\[ n = \frac{28}{2} \][/tex]
[tex]\[ n = 14 \][/tex]
So, there are 14 terms in the sequence.
## Summary of Answers:
1. [tex]\(a_1 = 31\)[/tex] and [tex]\(d = -7\)[/tex]
2. [tex]\(a_1 = 54\)[/tex]
3. The term at which the value is -18 is the 6th term.
4. There are 14 terms in the sequence.
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