Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Let's tackle each of the given problems one by one:
Problem 43:
What is the [tex]\(6^{\text{th}}\)[/tex] term of a sequence whose general term is [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex], where [tex]\(a_1 = 1\)[/tex] and [tex]\(a_2 = 3\)[/tex]?
Let's find the terms of the sequence step-by-step:
- [tex]\(a_1 = 1\)[/tex]
- [tex]\(a_2 = 3\)[/tex]
Now, using the given formula [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex]:
- [tex]\(a_3 = a_2 + 2 \cdot a_1 = 3 + 2 \cdot 1 = 3 + 2 = 4\)[/tex]
- [tex]\(a_4 = a_3 + 3 \cdot a_2 = 4 + 3 \cdot 3 = 4 + 9 = 13\)[/tex]
- [tex]\(a_5 = a_4 + 4 \cdot a_3 = 13 + 4 \cdot 4 = 13 + 16 = 29\)[/tex]
- [tex]\(a_6 = a_5 + 5 \cdot a_4 = 29 + 5 \cdot 13 = 29 + 65 = 94\)[/tex]
Therefore, the sixth term of the sequence is:
[tex]\[ \boxed{94} \][/tex]
The correct answer is none of the given options, indicating possibly a miscalculation or error in listing the correct choices.
Problem 44:
What is the general term [tex]\(g_n\)[/tex] of the geometric sequence whose first term is [tex]\(\frac{15}{2}\)[/tex] and third term is [tex]\(\frac{135}{8}\)[/tex]?
Let's find the common ratio [tex]\(r\)[/tex]:
We know:
- [tex]\(a_1 = \frac{15}{2}\)[/tex]
- [tex]\(a_3 = \frac{135}{8}\)[/tex]
In a geometric sequence, [tex]\(a_3 = a_1 \cdot r^2\)[/tex]. Using the given values:
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r^2 = \frac{135}{8} \times \frac{2}{15} = \frac{135}{60} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
Thus, the general term [tex]\(g_n\)[/tex] is:
[tex]\[ g_n = a_1 \cdot r^{n-1} = \frac{15}{2} \cdot \left(\frac{3}{2}\right)^{n-1} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5\left(\frac{3}{2}\right)^{n-1}} \][/tex]
Problem 45:
What is the [tex]\(15^{\text{th}}\)[/tex] partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
To find the partial sum, we first notice the pattern and identify whether it is an arithmetic sequence. We calculate the common difference [tex]\(d\)[/tex]:
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
[tex]\[ d = a_3 - a_2 = 8 - (-1) = 9 \][/tex]
Since this is an arithmetic sequence with:
- First term [tex]\(a = -10\)[/tex]
- Common difference [tex]\(d = 9\)[/tex]
The [tex]\(n^{\text{th}}\)[/tex] term of the sequence can be found by:
[tex]\[ a_n = a + (n-1)d = -10 + (n-1) \cdot 9 \][/tex]
The partial sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic sequence is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ S_{15} = \frac{15}{2} \left(2(-10) + (15-1) \cdot 9\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 = 15 \cdot 53 = 795 \][/tex]
Therefore, the [tex]\(15^{\text{th}}\)[/tex] partial sum is:
[tex]\[ \boxed{795} \][/tex]
Problem 46:
Which one of the following series is convergent?
Let's analyze each series:
A:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{\sqrt{5}}{2}\right)^n \][/tex]
This is a geometric series with [tex]\(a = \frac{\sqrt{5}}{2}\)[/tex] and [tex]\(r = \frac{\sqrt{5}}{2}\)[/tex]. Since [tex]\(\left|\frac{\sqrt{5}}{2}\right| > 1\)[/tex], the series diverges.
B:
[tex]\[ \sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n \][/tex]
This is an alternating series with [tex]\(|\text{ratio}| = \frac{4}{5} < 1\)[/tex]. It converges by the Alternating Series Test.
C:
[tex]\[ \sum_{n=1}^{\infty} 3^{n-3} \][/tex]
This can be rewritten as:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{27} \cdot 3^n \][/tex]
This is a geometric series with [tex]\(r = 3 > 1\)[/tex]. Therefore, it diverges.
D:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n+1}\left(\frac{3}{2}\right)^{n-1} \][/tex]
Combining the terms, we get:
[tex]\[ \left(\frac{4}{3} \cdot \frac{3}{2}\right)^n \cdot \frac{4}{3} \left(\frac{3}{2}\right)^{-1} = \left(\frac{4}{2}\right)^n \cdot \left(\frac{4}{3} \cdot \frac{2}{3}\right) = 2^n \cdot \frac{8}{9} \][/tex]
Thus, it grows exponentially and diverges.
Therefore, the correct answer is:
[tex]\[ \boxed{\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n} \][/tex]
Problem 47:
If the cost of a certain product in Birr as a function of time in days is given by [tex]\(f(t)=3t+t^2\)[/tex], then what is the average rate of change on the interval [tex]\(2 \leq t \leq 6\)[/tex]?
The average rate of change of a function [tex]\(f(t)\)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\(f(t) = 3t + t^2\)[/tex], with [tex]\(a = 2\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ f(6) = 3(6) + 6^2 = 18 + 36 = 54 \][/tex]
[tex]\[ f(2) = 3(2) + 2^2 = 6 + 4 = 10 \][/tex]
Therefore, the average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{11} \][/tex]
Problem 43:
What is the [tex]\(6^{\text{th}}\)[/tex] term of a sequence whose general term is [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex], where [tex]\(a_1 = 1\)[/tex] and [tex]\(a_2 = 3\)[/tex]?
Let's find the terms of the sequence step-by-step:
- [tex]\(a_1 = 1\)[/tex]
- [tex]\(a_2 = 3\)[/tex]
Now, using the given formula [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex]:
- [tex]\(a_3 = a_2 + 2 \cdot a_1 = 3 + 2 \cdot 1 = 3 + 2 = 4\)[/tex]
- [tex]\(a_4 = a_3 + 3 \cdot a_2 = 4 + 3 \cdot 3 = 4 + 9 = 13\)[/tex]
- [tex]\(a_5 = a_4 + 4 \cdot a_3 = 13 + 4 \cdot 4 = 13 + 16 = 29\)[/tex]
- [tex]\(a_6 = a_5 + 5 \cdot a_4 = 29 + 5 \cdot 13 = 29 + 65 = 94\)[/tex]
Therefore, the sixth term of the sequence is:
[tex]\[ \boxed{94} \][/tex]
The correct answer is none of the given options, indicating possibly a miscalculation or error in listing the correct choices.
Problem 44:
What is the general term [tex]\(g_n\)[/tex] of the geometric sequence whose first term is [tex]\(\frac{15}{2}\)[/tex] and third term is [tex]\(\frac{135}{8}\)[/tex]?
Let's find the common ratio [tex]\(r\)[/tex]:
We know:
- [tex]\(a_1 = \frac{15}{2}\)[/tex]
- [tex]\(a_3 = \frac{135}{8}\)[/tex]
In a geometric sequence, [tex]\(a_3 = a_1 \cdot r^2\)[/tex]. Using the given values:
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r^2 = \frac{135}{8} \times \frac{2}{15} = \frac{135}{60} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
Thus, the general term [tex]\(g_n\)[/tex] is:
[tex]\[ g_n = a_1 \cdot r^{n-1} = \frac{15}{2} \cdot \left(\frac{3}{2}\right)^{n-1} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5\left(\frac{3}{2}\right)^{n-1}} \][/tex]
Problem 45:
What is the [tex]\(15^{\text{th}}\)[/tex] partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
To find the partial sum, we first notice the pattern and identify whether it is an arithmetic sequence. We calculate the common difference [tex]\(d\)[/tex]:
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
[tex]\[ d = a_3 - a_2 = 8 - (-1) = 9 \][/tex]
Since this is an arithmetic sequence with:
- First term [tex]\(a = -10\)[/tex]
- Common difference [tex]\(d = 9\)[/tex]
The [tex]\(n^{\text{th}}\)[/tex] term of the sequence can be found by:
[tex]\[ a_n = a + (n-1)d = -10 + (n-1) \cdot 9 \][/tex]
The partial sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic sequence is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ S_{15} = \frac{15}{2} \left(2(-10) + (15-1) \cdot 9\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 = 15 \cdot 53 = 795 \][/tex]
Therefore, the [tex]\(15^{\text{th}}\)[/tex] partial sum is:
[tex]\[ \boxed{795} \][/tex]
Problem 46:
Which one of the following series is convergent?
Let's analyze each series:
A:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{\sqrt{5}}{2}\right)^n \][/tex]
This is a geometric series with [tex]\(a = \frac{\sqrt{5}}{2}\)[/tex] and [tex]\(r = \frac{\sqrt{5}}{2}\)[/tex]. Since [tex]\(\left|\frac{\sqrt{5}}{2}\right| > 1\)[/tex], the series diverges.
B:
[tex]\[ \sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n \][/tex]
This is an alternating series with [tex]\(|\text{ratio}| = \frac{4}{5} < 1\)[/tex]. It converges by the Alternating Series Test.
C:
[tex]\[ \sum_{n=1}^{\infty} 3^{n-3} \][/tex]
This can be rewritten as:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{27} \cdot 3^n \][/tex]
This is a geometric series with [tex]\(r = 3 > 1\)[/tex]. Therefore, it diverges.
D:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n+1}\left(\frac{3}{2}\right)^{n-1} \][/tex]
Combining the terms, we get:
[tex]\[ \left(\frac{4}{3} \cdot \frac{3}{2}\right)^n \cdot \frac{4}{3} \left(\frac{3}{2}\right)^{-1} = \left(\frac{4}{2}\right)^n \cdot \left(\frac{4}{3} \cdot \frac{2}{3}\right) = 2^n \cdot \frac{8}{9} \][/tex]
Thus, it grows exponentially and diverges.
Therefore, the correct answer is:
[tex]\[ \boxed{\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n} \][/tex]
Problem 47:
If the cost of a certain product in Birr as a function of time in days is given by [tex]\(f(t)=3t+t^2\)[/tex], then what is the average rate of change on the interval [tex]\(2 \leq t \leq 6\)[/tex]?
The average rate of change of a function [tex]\(f(t)\)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\(f(t) = 3t + t^2\)[/tex], with [tex]\(a = 2\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ f(6) = 3(6) + 6^2 = 18 + 36 = 54 \][/tex]
[tex]\[ f(2) = 3(2) + 2^2 = 6 + 4 = 10 \][/tex]
Therefore, the average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{11} \][/tex]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
