Answer :
Let's break down each part of the question and solve it step-by-step:
### 3.1 Given a geometric sequence: 6, 12, 24, 48, ...
#### 3.1.1 Determine the common ratio.
To find the common ratio ([tex]\( r \)[/tex]) of a geometric sequence, we divide a term by the preceding term:
[tex]\[ r = \frac{12}{6} = 2 \][/tex]
Hence, the common ratio is [tex]\( 2.0 \)[/tex].
#### 3.1.2 Determine the value of the first term.
The first term ([tex]\( a \)[/tex]) of the given sequence is the first number in the sequence:
[tex]\[ a = 6 \][/tex]
Therefore, the first term is [tex]\( 6 \)[/tex].
### 3.1.3 Determine [tex]\( T \)[/tex].
Given that if the numbers 42, 32, and 2 are added to the first, second, and third terms of a geometric sequence respectively, the three resulting terms will all be equal.
For a geometric sequence with first term [tex]\( a \)[/tex] and common ratio [tex]\( r \)[/tex]:
- The first term is [tex]\( a \)[/tex]
- The second term is [tex]\( ar \)[/tex]
- The third term is [tex]\( ar^2 \)[/tex]
Given:
[tex]\[ a + 42 = ar + 32 = ar^2 + 2 = T \][/tex]
Rewriting in terms of [tex]\( T \)[/tex]:
[tex]\[ T = a + 42 \][/tex]
[tex]\[ T = ar + 32 \][/tex]
[tex]\[ T = ar^2 + 2 \][/tex]
Next, solving for [tex]\( T \)[/tex]:
First term after adding 42: [tex]\( 6 + 42 = 48 \)[/tex]
Second term after adding 32: [tex]\( 12 + 32 = 44 \)[/tex]
Third term after adding 2: [tex]\( 24 + 2 = 26 \)[/tex]
To equalize these terms, we find their average:
[tex]\[ T = \frac{48 + 44 + 26}{3} = 39 \][/tex]
Thus, the value [tex]\( T \)[/tex] for the three equal terms is [tex]\( 39 \)[/tex].
### 3.3 The first term of an arithmetic sequence is 51 and the eighth term is 10.
#### 3.3.1 Determine the constant difference.
For an arithmetic sequence, the [tex]\( n \)[/tex]-th term can be defined as:
[tex]\[ a_n = a + (n-1)d \][/tex]
Here, the first term ([tex]\( a \)[/tex]) is 51, and the eighth term ([tex]\( a_8 \)[/tex]) is 10. Using the formula for the [tex]\( n \)[/tex]-th term:
[tex]\[ 10 = 51 + 7d \][/tex]
[tex]\[ 7d = 10 - 51 \][/tex]
[tex]\[ 7d = -41 \][/tex]
[tex]\[ d = \frac{-41}{7} = -5.857142857142857 \][/tex]
Therefore, the constant difference ([tex]\( d \)[/tex]) is approximately [tex]\( -5.857142857142857 \)[/tex].
#### 3.3.2 Find the twentieth term of the series.
Using the formula for the [tex]\( n \)[/tex]-th term again with [tex]\( n = 20 \)[/tex]:
[tex]\[ a_{20} = 51 + 19(-5.857142857142857) \][/tex]
[tex]\[ a_{20} = 51 - 111.14285714285714 \][/tex]
[tex]\[ a_{20} = -60.28571428571428 \][/tex]
Thus, the twentieth term is approximately [tex]\( -60.28571428571428 \)[/tex].
### 3.4 Calculate the sum of the multiples of 7 between 1 and 1000.
To find the sum of all multiples of 7 within the range (1 to 1000), we recognize that these form an arithmetic sequence with:
- First term ([tex]\( a \)[/tex]) = 7
- Common difference ([tex]\( d \)[/tex]) = 7
- Last term just under 1000 = 1000 (since [tex]\( 1001 = 1000 + 7 - 1 \)[/tex])
The number of terms ([tex]\( n \)[/tex]) can be found by setting up the equation for the last term of an arithmetic sequence:
[tex]\[ T_n = a + (n-1)d \][/tex]
[tex]\[ 1000 = 7 + (n-1)7 \][/tex]
[tex]\[ 1000 = 7n \][/tex]
[tex]\[ n = \frac{1000}{7} \approx 142.85714285714286 \rightarrow 142 \][/tex]
The sum ([tex]\( S \)[/tex]) of the first [tex]\( n \)[/tex] terms of an arithmetic sequence is given by:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
[tex]\[ S_{142} = \frac{142}{2} [2 \times 7 + (142-1) \times 7] \][/tex]
[tex]\[ S_{142} = 71 [14 + 141 \times 7] \][/tex]
[tex]\[ S_{142} = 71 [14 + 987] \][/tex]
[tex]\[ S_{142} = 71 \times 1001 \][/tex]
[tex]\[ S_{142} = 71071 \][/tex]
Thus, the sum of the multiples of 7 between 1 and 1000 is [tex]\( 71071 \)[/tex].
### 3.1 Given a geometric sequence: 6, 12, 24, 48, ...
#### 3.1.1 Determine the common ratio.
To find the common ratio ([tex]\( r \)[/tex]) of a geometric sequence, we divide a term by the preceding term:
[tex]\[ r = \frac{12}{6} = 2 \][/tex]
Hence, the common ratio is [tex]\( 2.0 \)[/tex].
#### 3.1.2 Determine the value of the first term.
The first term ([tex]\( a \)[/tex]) of the given sequence is the first number in the sequence:
[tex]\[ a = 6 \][/tex]
Therefore, the first term is [tex]\( 6 \)[/tex].
### 3.1.3 Determine [tex]\( T \)[/tex].
Given that if the numbers 42, 32, and 2 are added to the first, second, and third terms of a geometric sequence respectively, the three resulting terms will all be equal.
For a geometric sequence with first term [tex]\( a \)[/tex] and common ratio [tex]\( r \)[/tex]:
- The first term is [tex]\( a \)[/tex]
- The second term is [tex]\( ar \)[/tex]
- The third term is [tex]\( ar^2 \)[/tex]
Given:
[tex]\[ a + 42 = ar + 32 = ar^2 + 2 = T \][/tex]
Rewriting in terms of [tex]\( T \)[/tex]:
[tex]\[ T = a + 42 \][/tex]
[tex]\[ T = ar + 32 \][/tex]
[tex]\[ T = ar^2 + 2 \][/tex]
Next, solving for [tex]\( T \)[/tex]:
First term after adding 42: [tex]\( 6 + 42 = 48 \)[/tex]
Second term after adding 32: [tex]\( 12 + 32 = 44 \)[/tex]
Third term after adding 2: [tex]\( 24 + 2 = 26 \)[/tex]
To equalize these terms, we find their average:
[tex]\[ T = \frac{48 + 44 + 26}{3} = 39 \][/tex]
Thus, the value [tex]\( T \)[/tex] for the three equal terms is [tex]\( 39 \)[/tex].
### 3.3 The first term of an arithmetic sequence is 51 and the eighth term is 10.
#### 3.3.1 Determine the constant difference.
For an arithmetic sequence, the [tex]\( n \)[/tex]-th term can be defined as:
[tex]\[ a_n = a + (n-1)d \][/tex]
Here, the first term ([tex]\( a \)[/tex]) is 51, and the eighth term ([tex]\( a_8 \)[/tex]) is 10. Using the formula for the [tex]\( n \)[/tex]-th term:
[tex]\[ 10 = 51 + 7d \][/tex]
[tex]\[ 7d = 10 - 51 \][/tex]
[tex]\[ 7d = -41 \][/tex]
[tex]\[ d = \frac{-41}{7} = -5.857142857142857 \][/tex]
Therefore, the constant difference ([tex]\( d \)[/tex]) is approximately [tex]\( -5.857142857142857 \)[/tex].
#### 3.3.2 Find the twentieth term of the series.
Using the formula for the [tex]\( n \)[/tex]-th term again with [tex]\( n = 20 \)[/tex]:
[tex]\[ a_{20} = 51 + 19(-5.857142857142857) \][/tex]
[tex]\[ a_{20} = 51 - 111.14285714285714 \][/tex]
[tex]\[ a_{20} = -60.28571428571428 \][/tex]
Thus, the twentieth term is approximately [tex]\( -60.28571428571428 \)[/tex].
### 3.4 Calculate the sum of the multiples of 7 between 1 and 1000.
To find the sum of all multiples of 7 within the range (1 to 1000), we recognize that these form an arithmetic sequence with:
- First term ([tex]\( a \)[/tex]) = 7
- Common difference ([tex]\( d \)[/tex]) = 7
- Last term just under 1000 = 1000 (since [tex]\( 1001 = 1000 + 7 - 1 \)[/tex])
The number of terms ([tex]\( n \)[/tex]) can be found by setting up the equation for the last term of an arithmetic sequence:
[tex]\[ T_n = a + (n-1)d \][/tex]
[tex]\[ 1000 = 7 + (n-1)7 \][/tex]
[tex]\[ 1000 = 7n \][/tex]
[tex]\[ n = \frac{1000}{7} \approx 142.85714285714286 \rightarrow 142 \][/tex]
The sum ([tex]\( S \)[/tex]) of the first [tex]\( n \)[/tex] terms of an arithmetic sequence is given by:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
[tex]\[ S_{142} = \frac{142}{2} [2 \times 7 + (142-1) \times 7] \][/tex]
[tex]\[ S_{142} = 71 [14 + 141 \times 7] \][/tex]
[tex]\[ S_{142} = 71 [14 + 987] \][/tex]
[tex]\[ S_{142} = 71 \times 1001 \][/tex]
[tex]\[ S_{142} = 71071 \][/tex]
Thus, the sum of the multiples of 7 between 1 and 1000 is [tex]\( 71071 \)[/tex].
