Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
