Answer :
Sure! Let's break down the questions and solve them step-by-step.
### Kwv 5
Consider the following pattern:
[tex]\[ \begin{array}{ll} 1+2+3 & =6 \\ 4+5+6 & =15 \\ 7+8+9 & =24 \\ 10+11+12 & =33 \end{array} \][/tex]
Calculate the sum of the terms in the [tex]$2010^{\text{th}}$[/tex] row.
Each row contains three consecutive integers. Let's observe the pattern for the first four rows:
- The 1st row has numbers 1, 2, 3.
- The 2nd row has numbers 4, 5, 6.
- The 3rd row has numbers 7, 8, 9.
- The 4th row has numbers 10, 11, 12.
We can see that each row is an arithmetic sequence with three terms and a common difference of 1 between consecutive terms. The first number in the nth row can be calculated using the formula for the nth term of an arithmetic sequence.
The first term in the nth row [tex]\( S_n \)[/tex]:
[tex]\[ S_n = 1 + 3(n-1) \][/tex]
For the 2010th row:
[tex]\[ S_{2010} = 1 + 3(2010-1) = 1 + 3 \times 2009 = 1 + 6027 = 6028 \][/tex]
The terms in the 2010th row are:
[tex]\[ 6028, 6029, 6030 \][/tex]
The sum of these terms is:
[tex]\[ 6028 + 6029 + 6030 = 3 \times 6029 = 18087 \][/tex]
So, the sum of the terms in the 2010th row is [tex]\( 18087 \)[/tex].
### Kwv 6
Consider an arithmetic sequence which has the second term equal to 8 and [tex]\(T_5 = 10\)[/tex]:
a. Determine the first term and common difference of this sequence.
Let [tex]\( a \)[/tex] be the first term and [tex]\( d \)[/tex] be the common difference. In an arithmetic sequence, the nth term [tex]\( T_n \)[/tex] is given by:
[tex]\[ T_n = a + (n-1)d \][/tex]
For [tex]\( T_2 = 8 \)[/tex]:
[tex]\[ a + d = 8 \][/tex]
For [tex]\( T_5 = 10 \)[/tex]:
[tex]\[ a + 4d = 10 \][/tex]
We have two equations:
1. [tex]\( a + d = 8 \)[/tex]
2. [tex]\( a + 4d = 10 \)[/tex]
Subtracting the first equation from the second:
[tex]\[ (a + 4d) - (a + d) = 10 - 8 \][/tex]
[tex]\[ 3d = 2 \][/tex]
[tex]\[ d = \frac{2}{3} \][/tex]
Substitute [tex]\( d \)[/tex] back into the first equation:
[tex]\[ a + \frac{2}{3} = 8 \][/tex]
[tex]\[ a = 8 - \frac{2}{3} = \frac{24}{3} - \frac{2}{3} = \frac{22}{3} = 7.333... \][/tex]
So, the first term [tex]\( a \)[/tex] is [tex]\( 7.333... \)[/tex] and the common difference [tex]\( d \)[/tex] is [tex]\( 0.666... \)[/tex].
b. Determine the nth term [tex]\( T_n \)[/tex].
The nth term [tex]\( T_n \)[/tex] is given by:
[tex]\[ T_n = a + (n-1)d \][/tex]
[tex]\[ T_n = 7.333 + (n-1) \times 0.666 \][/tex]
c. Determine the sum of the first 50 terms.
The sum of the first n terms [tex]\( S_n \)[/tex] in an arithmetic sequence is given by:
[tex]\[ S_n = \frac{n}{2} (2a + (n-1)d) \][/tex]
For the first 50 terms:
[tex]\[ S_{50} = \frac{50}{2} \left[2 \times 7.333 + (50-1) \times 0.666\right] \][/tex]
[tex]\[ S_{50} = 25 \left[14.666 + 49 \times 0.666\right] \][/tex]
[tex]\[ S_{50} = 25 \left[14.666 + 32.666\right] \][/tex]
[tex]\[ S_{50} = 25 \times 47.333 = 1183.333 \][/tex]
So, the sum of the first 50 terms is [tex]\( 1183.333 \)[/tex].
The detailed answers are:
a. First term: [tex]\( 7.333... \)[/tex], common difference: [tex]\( 0.666... \)[/tex]
b. nth term: [tex]\( 7.333 + (n-1) \times 0.666 \)[/tex]
c. Sum of the first 50 terms: [tex]\( 1183.333 \)[/tex]
### Kwv 5
Consider the following pattern:
[tex]\[ \begin{array}{ll} 1+2+3 & =6 \\ 4+5+6 & =15 \\ 7+8+9 & =24 \\ 10+11+12 & =33 \end{array} \][/tex]
Calculate the sum of the terms in the [tex]$2010^{\text{th}}$[/tex] row.
Each row contains three consecutive integers. Let's observe the pattern for the first four rows:
- The 1st row has numbers 1, 2, 3.
- The 2nd row has numbers 4, 5, 6.
- The 3rd row has numbers 7, 8, 9.
- The 4th row has numbers 10, 11, 12.
We can see that each row is an arithmetic sequence with three terms and a common difference of 1 between consecutive terms. The first number in the nth row can be calculated using the formula for the nth term of an arithmetic sequence.
The first term in the nth row [tex]\( S_n \)[/tex]:
[tex]\[ S_n = 1 + 3(n-1) \][/tex]
For the 2010th row:
[tex]\[ S_{2010} = 1 + 3(2010-1) = 1 + 3 \times 2009 = 1 + 6027 = 6028 \][/tex]
The terms in the 2010th row are:
[tex]\[ 6028, 6029, 6030 \][/tex]
The sum of these terms is:
[tex]\[ 6028 + 6029 + 6030 = 3 \times 6029 = 18087 \][/tex]
So, the sum of the terms in the 2010th row is [tex]\( 18087 \)[/tex].
### Kwv 6
Consider an arithmetic sequence which has the second term equal to 8 and [tex]\(T_5 = 10\)[/tex]:
a. Determine the first term and common difference of this sequence.
Let [tex]\( a \)[/tex] be the first term and [tex]\( d \)[/tex] be the common difference. In an arithmetic sequence, the nth term [tex]\( T_n \)[/tex] is given by:
[tex]\[ T_n = a + (n-1)d \][/tex]
For [tex]\( T_2 = 8 \)[/tex]:
[tex]\[ a + d = 8 \][/tex]
For [tex]\( T_5 = 10 \)[/tex]:
[tex]\[ a + 4d = 10 \][/tex]
We have two equations:
1. [tex]\( a + d = 8 \)[/tex]
2. [tex]\( a + 4d = 10 \)[/tex]
Subtracting the first equation from the second:
[tex]\[ (a + 4d) - (a + d) = 10 - 8 \][/tex]
[tex]\[ 3d = 2 \][/tex]
[tex]\[ d = \frac{2}{3} \][/tex]
Substitute [tex]\( d \)[/tex] back into the first equation:
[tex]\[ a + \frac{2}{3} = 8 \][/tex]
[tex]\[ a = 8 - \frac{2}{3} = \frac{24}{3} - \frac{2}{3} = \frac{22}{3} = 7.333... \][/tex]
So, the first term [tex]\( a \)[/tex] is [tex]\( 7.333... \)[/tex] and the common difference [tex]\( d \)[/tex] is [tex]\( 0.666... \)[/tex].
b. Determine the nth term [tex]\( T_n \)[/tex].
The nth term [tex]\( T_n \)[/tex] is given by:
[tex]\[ T_n = a + (n-1)d \][/tex]
[tex]\[ T_n = 7.333 + (n-1) \times 0.666 \][/tex]
c. Determine the sum of the first 50 terms.
The sum of the first n terms [tex]\( S_n \)[/tex] in an arithmetic sequence is given by:
[tex]\[ S_n = \frac{n}{2} (2a + (n-1)d) \][/tex]
For the first 50 terms:
[tex]\[ S_{50} = \frac{50}{2} \left[2 \times 7.333 + (50-1) \times 0.666\right] \][/tex]
[tex]\[ S_{50} = 25 \left[14.666 + 49 \times 0.666\right] \][/tex]
[tex]\[ S_{50} = 25 \left[14.666 + 32.666\right] \][/tex]
[tex]\[ S_{50} = 25 \times 47.333 = 1183.333 \][/tex]
So, the sum of the first 50 terms is [tex]\( 1183.333 \)[/tex].
The detailed answers are:
a. First term: [tex]\( 7.333... \)[/tex], common difference: [tex]\( 0.666... \)[/tex]
b. nth term: [tex]\( 7.333 + (n-1) \times 0.666 \)[/tex]
c. Sum of the first 50 terms: [tex]\( 1183.333 \)[/tex]
