To solve for the 6th and 20th terms of the arithmetic sequence, we first need to identify the common difference and the general formula for the \(n\)-th term of the sequence.
The given sequence is:
- 1st term (\(a_1\)) = 50
- 2nd term (\(a_2\)) = 46
- 3rd term (\(a_3\)) = 42
- 4th term (\(a_4\)) = 38
We can calculate the common difference \(d\): \[ d = a_2 - a_1 = 46 - 50 = -4 \]
Now, we can express the \(n\)-th term of the sequence using the formula for the \(n\)-th term of an arithmetic sequence: \[ a_n = a_1 + (n - 1)d \]
Substituting \(a_1 = 50\) and \(d = -4\): \[ a_n = 50 + (n - 1)(-4) \] \[ a_n = 50 - 4(n - 1) \] \[ a_n = 50 - 4n + 4 \] \[ a_n = 54 - 4n \]
Now let's find the 6th term (\(a_6\)): \[ a_6 = 54 - 4(6) = 54 - 24 = 30 \]
Next, we find the 20th term (\(a_{20}\)): \[ a_{20} = 54 - 4(20) = 54 - 80 = -26 \]
Thus, the 6th term is 30 and the 20th term is -26.