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Question 43: What is the 6th 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], [tex]\( a_2 = 3 \)[/tex]?
To find the 6th term, we need to compute the terms up to [tex]\( a_6 \)[/tex] using the given recurrence relation:
1. [tex]\( a_1 = 1 \)[/tex]
2. [tex]\( a_2 = 3 \)[/tex]
3. [tex]\( a_3 = a_2 + (3-1)a_1 = 3 + 2 \times 1 = 5 \)[/tex]
4. [tex]\( a_4 = a_3 + (4-1)a_2 = 5 + 3 \times 3 = 14 \)[/tex]
5. [tex]\( a_5 = a_4 + (5-1)a_3 = 14 + 4 \times 5 = 34 \)[/tex]
6. [tex]\( a_6 = a_5 + (6-1)a_4 = 34 + 5 \times 14 = 34 + 70 = 104 \)[/tex]
So, the 6th term of the sequence [tex]\( a_n \)[/tex] is [tex]\( 62 \)[/tex].
Question 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]?
First, find the common ratio [tex]\( r \)[/tex]:
[tex]\[ a_1 = \frac{15}{2} \][/tex]
[tex]\[ a_3 = \frac{135}{8} \][/tex]
[tex]\[ a_3 = a_1 \cdot r^2 \][/tex]
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
[tex]\[ r^2 = \frac{\frac{135}{8}}{\frac{15}{2}} = \frac{135}{8} \times \frac{2}{15} = \frac{135 \times 2}{8 \times 15} = \frac{270}{120} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
The general term [tex]\( g_n \)[/tex] is:
[tex]\[ g_n = 5 \left( \frac{3}{2} \right)^{n-1} \][/tex]
Question 45: What is the 15th partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
This sequence is an arithmetic sequence with:
[tex]\[ a_1 = -10 \][/tex]
[tex]\[ a_2 = -1 \][/tex]
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
The formula for the sum 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]\[ S_{15} = \frac{15}{2} \left(2 \cdot (-10) + (15-1) \cdot 9 \right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 \][/tex]
[tex]\[ S_{15} = 15 \cdot 53 = 795 \][/tex]
Question 46: Which one of the following series is convergent?
A. [tex]\( \sum_{n=1}^{\infty} \left( \frac{\sqrt{5}}{2} \right)^n \)[/tex]
B. [tex]\( \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{4}{5} \right)^n \)[/tex]
C. [tex]\( \sum_{n=1}^{\infty} 3^{n-3} \)[/tex]
D. [tex]\( \sum_{n=1}^{\infty} \left( \frac{4}{3} \right)^{n+1} \left( \frac{3}{2} \right)^{n-1} \)[/tex]
Series A converges because the common ratio [tex]\( \frac{\sqrt{5}}{2} < 1 \)[/tex].
Series B converges because it is an alternating series with decreasing terms in absolute value.
Series C diverges because the common ratio [tex]\( 3 > 1 \)[/tex].
Series D diverges because the common ratio [tex]\( \left( \frac{4}{3} \right)^{n+1} \left( \frac{3}{2} \right)^{n-1} > 1 \)[/tex] eventually.
So, the convergent series is A.
Question 47: If the cost of a certain product is given by [tex]\( f(t) = 3t + t^2 \)[/tex], 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 the interval [tex]\( [a, b] \)[/tex] is calculated as:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\( f(t) = 3t + t^2 \)[/tex], over the interval [tex]\( [2, 6] \)[/tex]:
[tex]\[ f(2) = 3 \cdot 2 + 2^2 = 6 + 4 = 10 \][/tex]
[tex]\[ f(6) = 3 \cdot 6 + 6^2 = 18 + 36 = 54 \][/tex]
The average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{f(6) - f(2)}{6 - 2} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
So, the average rate of change is [tex]\( 11 \)[/tex].
Question 43: What is the 6th 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], [tex]\( a_2 = 3 \)[/tex]?
To find the 6th term, we need to compute the terms up to [tex]\( a_6 \)[/tex] using the given recurrence relation:
1. [tex]\( a_1 = 1 \)[/tex]
2. [tex]\( a_2 = 3 \)[/tex]
3. [tex]\( a_3 = a_2 + (3-1)a_1 = 3 + 2 \times 1 = 5 \)[/tex]
4. [tex]\( a_4 = a_3 + (4-1)a_2 = 5 + 3 \times 3 = 14 \)[/tex]
5. [tex]\( a_5 = a_4 + (5-1)a_3 = 14 + 4 \times 5 = 34 \)[/tex]
6. [tex]\( a_6 = a_5 + (6-1)a_4 = 34 + 5 \times 14 = 34 + 70 = 104 \)[/tex]
So, the 6th term of the sequence [tex]\( a_n \)[/tex] is [tex]\( 62 \)[/tex].
Question 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]?
First, find the common ratio [tex]\( r \)[/tex]:
[tex]\[ a_1 = \frac{15}{2} \][/tex]
[tex]\[ a_3 = \frac{135}{8} \][/tex]
[tex]\[ a_3 = a_1 \cdot r^2 \][/tex]
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
[tex]\[ r^2 = \frac{\frac{135}{8}}{\frac{15}{2}} = \frac{135}{8} \times \frac{2}{15} = \frac{135 \times 2}{8 \times 15} = \frac{270}{120} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
The general term [tex]\( g_n \)[/tex] is:
[tex]\[ g_n = 5 \left( \frac{3}{2} \right)^{n-1} \][/tex]
Question 45: What is the 15th partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
This sequence is an arithmetic sequence with:
[tex]\[ a_1 = -10 \][/tex]
[tex]\[ a_2 = -1 \][/tex]
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
The formula for the sum 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]\[ S_{15} = \frac{15}{2} \left(2 \cdot (-10) + (15-1) \cdot 9 \right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 \][/tex]
[tex]\[ S_{15} = 15 \cdot 53 = 795 \][/tex]
Question 46: Which one of the following series is convergent?
A. [tex]\( \sum_{n=1}^{\infty} \left( \frac{\sqrt{5}}{2} \right)^n \)[/tex]
B. [tex]\( \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{4}{5} \right)^n \)[/tex]
C. [tex]\( \sum_{n=1}^{\infty} 3^{n-3} \)[/tex]
D. [tex]\( \sum_{n=1}^{\infty} \left( \frac{4}{3} \right)^{n+1} \left( \frac{3}{2} \right)^{n-1} \)[/tex]
Series A converges because the common ratio [tex]\( \frac{\sqrt{5}}{2} < 1 \)[/tex].
Series B converges because it is an alternating series with decreasing terms in absolute value.
Series C diverges because the common ratio [tex]\( 3 > 1 \)[/tex].
Series D diverges because the common ratio [tex]\( \left( \frac{4}{3} \right)^{n+1} \left( \frac{3}{2} \right)^{n-1} > 1 \)[/tex] eventually.
So, the convergent series is A.
Question 47: If the cost of a certain product is given by [tex]\( f(t) = 3t + t^2 \)[/tex], 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 the interval [tex]\( [a, b] \)[/tex] is calculated as:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\( f(t) = 3t + t^2 \)[/tex], over the interval [tex]\( [2, 6] \)[/tex]:
[tex]\[ f(2) = 3 \cdot 2 + 2^2 = 6 + 4 = 10 \][/tex]
[tex]\[ f(6) = 3 \cdot 6 + 6^2 = 18 + 36 = 54 \][/tex]
The average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{f(6) - f(2)}{6 - 2} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
So, the average rate of change is [tex]\( 11 \)[/tex].
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