Answer :
Sure, let's tackle each part of the question step-by-step.
### Part (a): Define Arithmetic Series
1. Arithmetic Series
An arithmetic series is a sequence of numbers in which the difference of any two successive members is a constant. This constant difference is known as the common difference, denoted as [tex]\( d \)[/tex]. For example, in the sequence [tex]\( a, a+d, a+2d, \ldots \)[/tex], [tex]\( a \)[/tex] is the first term, and each term increases by [tex]\( d \)[/tex].
### Part (b): Find the First Term and Common Difference
2. Given:
- Number of terms, [tex]\( n = 14 \)[/tex]
- Sum of the first 14 terms, [tex]\( S_{14} = 420 \)[/tex]
- Last term is 4 times the second term.
3. Sum of the first [tex]\( n \)[/tex] terms of an AP:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \][/tex]
Substituting given values:
[tex]\[ 420 = \frac{14}{2} \times (2a + 13d) \][/tex]
Simplifying:
[tex]\[ 420 = 7 \times (2a + 13d) \][/tex]
[tex]\[ 60 = 2a + 13d \][/tex]
4. Relation between the last term and the second term:
The last term can be expressed as:
[tex]\[ a_{14} = a + 13d \][/tex]
It is also given that:
[tex]\[ a_{14} = 4(a + d) \][/tex]
Thus,
[tex]\[ a + 13d = 4(a + d) \][/tex]
Simplifying:
[tex]\[ a + 13d = 4a + 4d \][/tex]
[tex]\[ a = 3d \][/tex]
5. Substitute [tex]\( a = 3d \)[/tex] into the sum equation:
[tex]\[ 60 = 2(3d) + 13d \][/tex]
[tex]\[ 60 = 6d + 13d \][/tex]
[tex]\[ 60 = 19d \][/tex]
Solving for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{60}{19} \approx 3.1578947368421053 \][/tex]
6. Calculate the first term [tex]\( a \)[/tex]:
[tex]\[ a = 3d \][/tex]
[tex]\[ a = 3 \times 3.1578947368421053 \approx 9.473684210526315 \][/tex]
### Part (c): Find the Sum of the First 20 Terms
7. Sum of the first 20 terms:
[tex]\[ S_{20} = \frac{20}{2} \times (2a + 19d) \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ S_{20} = 10 \times (2 \times 9.473684210526315 + 19 \times 3.1578947368421053) \][/tex]
[tex]\[ S_{20} = 10 \times (18.94736842105263 + 59.99999999999999) \][/tex]
[tex]\[ S_{20} = 10 \times 78.94736842105262 \][/tex]
[tex]\[ S_{20} = 789.4736842105262 \][/tex]
### Summary:
(a) An arithmetic series is a sequence of numbers where the difference between successive terms is constant.
(b) The first term [tex]\( a \)[/tex] is approximately 9.473684210526315, and the common difference [tex]\( d \)[/tex] is approximately 3.1578947368421053.
(c) The sum of the first 20 terms is approximately 789.4736842105262.
### Part (a): Define Arithmetic Series
1. Arithmetic Series
An arithmetic series is a sequence of numbers in which the difference of any two successive members is a constant. This constant difference is known as the common difference, denoted as [tex]\( d \)[/tex]. For example, in the sequence [tex]\( a, a+d, a+2d, \ldots \)[/tex], [tex]\( a \)[/tex] is the first term, and each term increases by [tex]\( d \)[/tex].
### Part (b): Find the First Term and Common Difference
2. Given:
- Number of terms, [tex]\( n = 14 \)[/tex]
- Sum of the first 14 terms, [tex]\( S_{14} = 420 \)[/tex]
- Last term is 4 times the second term.
3. Sum of the first [tex]\( n \)[/tex] terms of an AP:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \][/tex]
Substituting given values:
[tex]\[ 420 = \frac{14}{2} \times (2a + 13d) \][/tex]
Simplifying:
[tex]\[ 420 = 7 \times (2a + 13d) \][/tex]
[tex]\[ 60 = 2a + 13d \][/tex]
4. Relation between the last term and the second term:
The last term can be expressed as:
[tex]\[ a_{14} = a + 13d \][/tex]
It is also given that:
[tex]\[ a_{14} = 4(a + d) \][/tex]
Thus,
[tex]\[ a + 13d = 4(a + d) \][/tex]
Simplifying:
[tex]\[ a + 13d = 4a + 4d \][/tex]
[tex]\[ a = 3d \][/tex]
5. Substitute [tex]\( a = 3d \)[/tex] into the sum equation:
[tex]\[ 60 = 2(3d) + 13d \][/tex]
[tex]\[ 60 = 6d + 13d \][/tex]
[tex]\[ 60 = 19d \][/tex]
Solving for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{60}{19} \approx 3.1578947368421053 \][/tex]
6. Calculate the first term [tex]\( a \)[/tex]:
[tex]\[ a = 3d \][/tex]
[tex]\[ a = 3 \times 3.1578947368421053 \approx 9.473684210526315 \][/tex]
### Part (c): Find the Sum of the First 20 Terms
7. Sum of the first 20 terms:
[tex]\[ S_{20} = \frac{20}{2} \times (2a + 19d) \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ S_{20} = 10 \times (2 \times 9.473684210526315 + 19 \times 3.1578947368421053) \][/tex]
[tex]\[ S_{20} = 10 \times (18.94736842105263 + 59.99999999999999) \][/tex]
[tex]\[ S_{20} = 10 \times 78.94736842105262 \][/tex]
[tex]\[ S_{20} = 789.4736842105262 \][/tex]
### Summary:
(a) An arithmetic series is a sequence of numbers where the difference between successive terms is constant.
(b) The first term [tex]\( a \)[/tex] is approximately 9.473684210526315, and the common difference [tex]\( d \)[/tex] is approximately 3.1578947368421053.
(c) The sum of the first 20 terms is approximately 789.4736842105262.
