Answer :
Sure, let's solve this problem step-by-step.
### Part (a): Difference between Arithmetic Series and Geometric Series
- Arithmetic Series: In an arithmetic series, each term is obtained by adding a constant difference to the previous term. If [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference, then the terms of the series are:
[tex]\[ a, a + d, a + 2d, a + 3d, \ldots \][/tex]
- Geometric Series: In a geometric series, each term is obtained by multiplying the previous term by a constant ratio. If [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio, then the terms of the series are:
[tex]\[ a, ar, ar^2, ar^3, \ldots \][/tex]
### Part (b): Find the First Term of the Arithmetic Series
Let's denote the three terms of the arithmetic series by [tex]\( a \)[/tex], [tex]\( a + d \)[/tex], and [tex]\( a + 2d \)[/tex].
Given: The sum of these three terms is 24.
[tex]\[ a + (a + d) + (a + 2d) = 24 \][/tex]
[tex]\[ 3a + 3d = 24 \][/tex]
[tex]\[ a + d = 8 \quad \text{(1)} \][/tex]
### Part (c): Find the Positive Common Difference
Given: Adding 1, 6, and 18 to the terms forms a geometric series: [tex]\( a + 1 \)[/tex], [tex]\( a + d + 6 \)[/tex], [tex]\( a + 2d + 18 \)[/tex].
For these to be in a geometric series, the ratio between consecutive terms must be constant:
[tex]\[ \frac{a + d + 6}{a + 1} = \frac{a + 2d + 18}{a + d + 6} \quad \text{(2)} \][/tex]
From equation (1), we know:
[tex]\[ a + d = 8 \quad \Rightarrow \quad d = 8 - a \quad \text{(3)} \][/tex]
Using [tex]\( a + d = 8 \)[/tex], substitute [tex]\( a + d \)[/tex] by 8 in equation (2):
[tex]\[ \frac{8 + 6}{a + 1} = \frac{16 + 18 - a}{8 + 6} \][/tex]
[tex]\[ \frac{14}{a + 1} = \frac{34 - a}{14} \][/tex]
Cross multiply:
[tex]\[ 14^2 = (a + 1)(34 - a) \][/tex]
[tex]\[ 196 = 34a - a^2 + 34 - a \][/tex]
[tex]\[ 196 = 34 + 33a - a^2 \][/tex]
Solve the quadratic equation:
[tex]\[ a^2 - 33a + 162 = 0 \][/tex]
Using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = \frac{33 \pm \sqrt{33^2 - 4 \cdot 1 \cdot 162}}{2 \cdot 1} \][/tex]
[tex]\[ a = \frac{33 \pm \sqrt{1089 - 648}}{2} \][/tex]
[tex]\[ a = \frac{33 \pm \sqrt{441}}{2} \][/tex]
[tex]\[ a = \frac{33 \pm 21}{2} \][/tex]
So, the possible values for [tex]\( a \)[/tex] are:
[tex]\[ a = \frac{33 + 21}{2} = 27 \quad \text{and} \quad a = \frac{33 - 21}{2} = 6 \][/tex]
We discard [tex]\( a = 27 \)[/tex] as it leads to a negative common difference. Hence:
[tex]\[ a = 6 \][/tex]
Using [tex]\( a + d = 8 \)[/tex]:
[tex]\[ 6 + d = 8 \quad \Rightarrow \quad d = 2 \][/tex]
### Part (d): Which Term of the Arithmetic Series is 104?
The general term of an arithmetic series is given by [tex]\( a_n = a + (n-1)d \)[/tex].
Given [tex]\( a = 6 \)[/tex] and [tex]\( d = 2 \)[/tex]:
[tex]\[ 104 = 6 + (n-1) \cdot 2 \][/tex]
[tex]\[ 104 = 6 + 2n - 2 \][/tex]
[tex]\[ 104 = 2n + 4 \][/tex]
[tex]\[ 100 = 2n \][/tex]
[tex]\[ n = 50 \][/tex]
So, the 50th term of the arithmetic series is 104.
### Part (e): Compare the Sum of First Five Terms of Arithmetic Series and First Four Terms of Geometric Series
1. Sum of First Five Terms of Arithmetic Series: [tex]\( S_5 \)[/tex]
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic series is given by:
[tex]\[ S_n = \frac{n}{2}(2a + (n-1)d) \][/tex]
For [tex]\( n = 5 \)[/tex], [tex]\( a = 6 \)[/tex], and [tex]\( d = 2 \)[/tex]:
[tex]\[ S_5 = \frac{5}{2}(2 \cdot 6 + (5-1) \cdot 2) \][/tex]
[tex]\[ S_5 = \frac{5}{2}(12 + 8) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \cdot 20 \][/tex]
[tex]\[ S_5 = 50 \][/tex]
2. Sum of First Four Terms of Geometric Series: [tex]\( S_4 \)[/tex]
The first term of the geometric series is [tex]\( a + 1 = 6 + 1 = 7 \)[/tex].
The common ratio [tex]\( r \)[/tex] is given by:
[tex]\[ r = \frac{a + d + 6}{a + 1} = \frac{8}{7} \][/tex]
The sum [tex]\( S_n \)[/tex] for the first [tex]\( n \)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \frac{r^n-1}{r-1} \][/tex]
For [tex]\( n = 4 \)[/tex], [tex]\( a = 7 \)[/tex], and [tex]\( r = \frac{8}{7} \)[/tex]:
[tex]\[ S_4 = 7 \frac{(\frac{8}{7})^4 - 1}{\frac{8}{7} - 1} \][/tex]
[tex]\[ S_4 = 7 \frac{\left( \frac{4096}{2401} \right) - 1}{\frac{1}{7}} \][/tex]
[tex]\[ S_4 = 7 \cdot 7 \left( \frac{4096 - 2401}{2401} \right) \][/tex]
[tex]\[ S_4 = 49 \left( \frac{1695}{2401} \right) \][/tex]
[tex]\[ S_4 = \frac{83055}{2401} \approx 34.6 \][/tex]
Comparison:
[tex]\[ S_5 (Arithmetic Series) = 50 > S_4 (Geometric Series) = 34.6 \][/tex]
Thus, the sum of the first five terms of the arithmetic series is greater than the sum of the first four terms of the geometric series.
### Part (a): Difference between Arithmetic Series and Geometric Series
- Arithmetic Series: In an arithmetic series, each term is obtained by adding a constant difference to the previous term. If [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference, then the terms of the series are:
[tex]\[ a, a + d, a + 2d, a + 3d, \ldots \][/tex]
- Geometric Series: In a geometric series, each term is obtained by multiplying the previous term by a constant ratio. If [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio, then the terms of the series are:
[tex]\[ a, ar, ar^2, ar^3, \ldots \][/tex]
### Part (b): Find the First Term of the Arithmetic Series
Let's denote the three terms of the arithmetic series by [tex]\( a \)[/tex], [tex]\( a + d \)[/tex], and [tex]\( a + 2d \)[/tex].
Given: The sum of these three terms is 24.
[tex]\[ a + (a + d) + (a + 2d) = 24 \][/tex]
[tex]\[ 3a + 3d = 24 \][/tex]
[tex]\[ a + d = 8 \quad \text{(1)} \][/tex]
### Part (c): Find the Positive Common Difference
Given: Adding 1, 6, and 18 to the terms forms a geometric series: [tex]\( a + 1 \)[/tex], [tex]\( a + d + 6 \)[/tex], [tex]\( a + 2d + 18 \)[/tex].
For these to be in a geometric series, the ratio between consecutive terms must be constant:
[tex]\[ \frac{a + d + 6}{a + 1} = \frac{a + 2d + 18}{a + d + 6} \quad \text{(2)} \][/tex]
From equation (1), we know:
[tex]\[ a + d = 8 \quad \Rightarrow \quad d = 8 - a \quad \text{(3)} \][/tex]
Using [tex]\( a + d = 8 \)[/tex], substitute [tex]\( a + d \)[/tex] by 8 in equation (2):
[tex]\[ \frac{8 + 6}{a + 1} = \frac{16 + 18 - a}{8 + 6} \][/tex]
[tex]\[ \frac{14}{a + 1} = \frac{34 - a}{14} \][/tex]
Cross multiply:
[tex]\[ 14^2 = (a + 1)(34 - a) \][/tex]
[tex]\[ 196 = 34a - a^2 + 34 - a \][/tex]
[tex]\[ 196 = 34 + 33a - a^2 \][/tex]
Solve the quadratic equation:
[tex]\[ a^2 - 33a + 162 = 0 \][/tex]
Using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = \frac{33 \pm \sqrt{33^2 - 4 \cdot 1 \cdot 162}}{2 \cdot 1} \][/tex]
[tex]\[ a = \frac{33 \pm \sqrt{1089 - 648}}{2} \][/tex]
[tex]\[ a = \frac{33 \pm \sqrt{441}}{2} \][/tex]
[tex]\[ a = \frac{33 \pm 21}{2} \][/tex]
So, the possible values for [tex]\( a \)[/tex] are:
[tex]\[ a = \frac{33 + 21}{2} = 27 \quad \text{and} \quad a = \frac{33 - 21}{2} = 6 \][/tex]
We discard [tex]\( a = 27 \)[/tex] as it leads to a negative common difference. Hence:
[tex]\[ a = 6 \][/tex]
Using [tex]\( a + d = 8 \)[/tex]:
[tex]\[ 6 + d = 8 \quad \Rightarrow \quad d = 2 \][/tex]
### Part (d): Which Term of the Arithmetic Series is 104?
The general term of an arithmetic series is given by [tex]\( a_n = a + (n-1)d \)[/tex].
Given [tex]\( a = 6 \)[/tex] and [tex]\( d = 2 \)[/tex]:
[tex]\[ 104 = 6 + (n-1) \cdot 2 \][/tex]
[tex]\[ 104 = 6 + 2n - 2 \][/tex]
[tex]\[ 104 = 2n + 4 \][/tex]
[tex]\[ 100 = 2n \][/tex]
[tex]\[ n = 50 \][/tex]
So, the 50th term of the arithmetic series is 104.
### Part (e): Compare the Sum of First Five Terms of Arithmetic Series and First Four Terms of Geometric Series
1. Sum of First Five Terms of Arithmetic Series: [tex]\( S_5 \)[/tex]
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic series is given by:
[tex]\[ S_n = \frac{n}{2}(2a + (n-1)d) \][/tex]
For [tex]\( n = 5 \)[/tex], [tex]\( a = 6 \)[/tex], and [tex]\( d = 2 \)[/tex]:
[tex]\[ S_5 = \frac{5}{2}(2 \cdot 6 + (5-1) \cdot 2) \][/tex]
[tex]\[ S_5 = \frac{5}{2}(12 + 8) \][/tex]
[tex]\[ S_5 = \frac{5}{2} \cdot 20 \][/tex]
[tex]\[ S_5 = 50 \][/tex]
2. Sum of First Four Terms of Geometric Series: [tex]\( S_4 \)[/tex]
The first term of the geometric series is [tex]\( a + 1 = 6 + 1 = 7 \)[/tex].
The common ratio [tex]\( r \)[/tex] is given by:
[tex]\[ r = \frac{a + d + 6}{a + 1} = \frac{8}{7} \][/tex]
The sum [tex]\( S_n \)[/tex] for the first [tex]\( n \)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \frac{r^n-1}{r-1} \][/tex]
For [tex]\( n = 4 \)[/tex], [tex]\( a = 7 \)[/tex], and [tex]\( r = \frac{8}{7} \)[/tex]:
[tex]\[ S_4 = 7 \frac{(\frac{8}{7})^4 - 1}{\frac{8}{7} - 1} \][/tex]
[tex]\[ S_4 = 7 \frac{\left( \frac{4096}{2401} \right) - 1}{\frac{1}{7}} \][/tex]
[tex]\[ S_4 = 7 \cdot 7 \left( \frac{4096 - 2401}{2401} \right) \][/tex]
[tex]\[ S_4 = 49 \left( \frac{1695}{2401} \right) \][/tex]
[tex]\[ S_4 = \frac{83055}{2401} \approx 34.6 \][/tex]
Comparison:
[tex]\[ S_5 (Arithmetic Series) = 50 > S_4 (Geometric Series) = 34.6 \][/tex]
Thus, the sum of the first five terms of the arithmetic series is greater than the sum of the first four terms of the geometric series.
