1) Given the arithmetic sequence an = 4 - 3(n - 1), what is the domain for n?
All integers where n ≥ 1
All integers where n > 1
All integers where n ≤ 4
All integers where n ≥ 4
2) What is the 6th term of the geometric sequence where a1 = 1,024 and a4 = -16?
1
-0.25
-1
0.25
1) To find the domain for n in the arithmetic sequence an = 4 - 3(n - 1), we need to consider the values of n that make sense in the context of the sequence.
In an arithmetic sequence, the subscript "n" represents the position of a term in the sequence. The domain for n depends on how many terms are in the given sequence.
In this case, the given formula is an = 4 - 3(n - 1), which suggests that the first term of the sequence is when n = 1. Therefore, n should be a positive integer or zero if we assume the sequence starts from zero.
Since there are no restrictions mentioned in the problem, the domain for n would be all integers where n ≥ 1.
Thus, the correct answer is: All integers where n ≥ 1.
2) To find the 6th term of a geometric sequence, we need to determine the common ratio (r) of the sequence and apply the formula an = a1 * r^(n-1), where a1 represents the first term of the sequence.
In this case, we are given two terms of the geometric sequence: a1 = 1,024 and a4 = -16.
By comparing these two terms, we can determine the common ratio:
a4 = a1 * r^(4-1)
-16 = 1,024 * r^3
To find r, we can solve this equation:
r^3 = -16/1,024
r^3 = -1/64
Taking the cube root of both sides:
r = -1/4
Now that we have the common ratio, we can find the 6th term:
a6 = a1 * r^(6-1)
a6 = 1,024 * (-1/4)^5
a6 = 1,024 * -1/1024
a6 = -1
Thus, the 6th term of the given geometric sequence is -1.
Therefore, the correct answer is: -1.
1) To find the domain for n in the given arithmetic sequence an = 4 - 3(n - 1), we need to consider the values of n that would make sense in the context of the sequence.
The general form of an arithmetic sequence is an = a1 + (n - 1)d, where a1 is the first term and d is the common difference. In this case, a1 = 4 and d = -3.
We can find the domain by finding the range of n that would give us valid terms in the sequence.
Since the first term corresponds to n = 1, plugging n = 1 into the formula gives us a1 = 4 - 3(1 - 1) = 4.
Now, let's find the value of n that gives us the last term of the sequence.
The last term is given by an = 4 - 3(n - 1). If we let an = 0, we can solve for n:
0 = 4 - 3(n - 1)
0 = 4 - 3n + 3
-7 = -3n
n = -7/-3 = 7/3
Since we cannot have a fraction as n in this context, we'll take the integer part of 7/3, which is 2.
Therefore, the valid terms for n in the given arithmetic sequence are n = 1, 2.
So, the correct answer is: All integers where n ≥ 1.
2) The geometric sequence is a sequence in which each term is found by multiplying the previous term by a constant factor, called the common ratio, denoted as r.
We are given a1 = 1024 and a4 = -16.
The general form of a geometric sequence is an = a1 * r^(n-1).
From the given information, we can set up two equations:
a4 = a1 * r^(4-1) = -16, and
a1 = 1024.
Substitute a1 = 1024 into the first equation:
1024 * r^3 = -16.
Divide both sides by 1024:
r^3 = -16/1024 = -1/64.
Cube root both sides to solve for r:
r = -1/4.
Now, we can find the 6th term of the sequence using a1 = 1024 and r = -1/4:
a6 = a1 * r^(6-1) = 1024 * (-1/4)^(5) = 1024 * (-1/1024) = -1.
Therefore, the 6th term of the given geometric sequence is -1.
So, the correct answer is: -1.