Let's solve each question step by step.
Question 1:
Given: \( a_n = 4 - 3n \)
To find the 7th term, we calculate \( a_7 \): \[ a_7 = 4 - 3(7) = 4 - 21 = -17 \] Answer: \( a_7 = -17 \)
Question 2:
Given: \( a_1 = 5 \) and \( a_n = 25a_{n-1} \)
To find the second term \( a_2 \): \[ a_2 = 25a_1 = 25 \times 5 = 125 \] Answer: \( a_2 = 125 \)
Question 3:
Given: \( a_1 = 56 \) and \( a_n = 4a_{n-1} \)
Here, the common ratio \( r \) is simply the multiplier we use to go from one term to the next: \[ r = 4 \] Answer: \( r = 4 \)
Question 4:
The sequence is: 7, 5, 3, 1, -1...
To determine the recursive formula, we notice that the difference between consecutive terms is -2:
- From 7 to 5: \(-2\)
- From 5 to 3: \(-2\)
- From 3 to 1: \(-2\)
- From 1 to -1: \(-2\)
Thus, the recursive formula is: \[ a_1 = 7, \quad a_n = a_{n-1} - 2 \] Answer: \( a_1 = 7, a_n = a_{n-1} - 2 \)
Question 5:
Given: \( A_n = 14 + (n - 1) \cdot 12 \)
To find the 9th term: \[ A_9 = 14 + (9 - 1) \cdot 12 = 14 + (8) \cdot 12 = 14 + 96 = 110 \] Answer: \( A_9 = 110 \)
As a recap the answers to each respective question are:
- \( a_7 = -17 \)
- \( a_2 = 125 \)
- Common ratio \( r = 4 \)
- \( a_1 = 7, a_n = a_{n-1} - 2 \)
- \( A_9 = 110 \)