To analyze both sequences and determine the values of \( a_7 \), we first need to identify the pattern in each sequence.
Problem 4
Given the sequence: 19, 15, 11, 7, 3...
This sequence is a linear sequence where each term decreases by 4. We can express this as:
- \( a_1 = 19 \)
- \( a_2 = 15 = 19 - 4 \)
- \( a_3 = 11 = 19 - 8 \)
- \( a_4 = 7 = 19 - 12 \)
- \( a_5 = 3 = 19 - 16 \)
The general term can be described as: \[ a_n = 19 - 4(n - 1) \] Alternatively, this can be simplified to: \[ a_n = 23 - 4n \]
Now, we can find \( a_7 \): \[ a_7 = 23 - 4(7) = 23 - 28 = -5 \]
Thus, the answer for problem 4 is: B) a7 = −5
Problem 5
Given the sequence: 81, 27, 9, 3, 1...
This sequence is a geometric sequence where each term is obtained by multiplying the previous term by \(\frac{1}{3}\). We can express this as:
- \( a_1 = 81 \)
- \( a_2 = 27 = 81 \times \frac{1}{3} \)
- \( a_3 = 9 = 27 \times \frac{1}{3} \)
- \( a_4 = 3 = 9 \times \frac{1}{3} \)
- \( a_5 = 1 = 3 \times \frac{1}{3} \)
The general term can be described as: \[ a_n = 81 \times \left(\frac{1}{3}\right)^{n-1} \]
Now, we can find \( a_7 \): \[ a_7 = 81 \times \left(\frac{1}{3}\right)^{7-1} = 81 \times \left(\frac{1}{3}\right)^{6} = 81 \times \frac{1}{729} \]
Since \( 81 = 3^4 \) and \( 729 = 3^6 \): \[ a_7 = \frac{3^4}{3^6} = \frac{1}{3^{6-4}} = \frac{1}{3^2} = \frac{1}{9} \]
Thus, the answer for problem 5 is: C) a7 = 1/9
4 answers