Final answer:
To find the maximum sum of the sequence, the first term is calculated as 32 and the common difference as -4. As the terms decrease, to maximize the sum, we consider only positive terms, resulting in a sum of the first 8 terms, which is 144.
Step-by-step explanation:
Finding the Maximum Sum of a Sequence
To find the maximum sum of a sequence whose second term is twenty-eight and fourth term is twenty, we need to recognize that the sequence being referred to is likely an arithmetic sequence. This type of sequence has a constant difference between consecutive terms. We'll need to find this common difference and the first term to determine the maximum sum.
Let's designate the first term of the sequence as a and the common difference as d. The second term then is a + d, which we know is equal to 28, and the fourth term is a + 3d, which equals 20. Now we have two equations:
a + d = 28
a + 3d = 20
Subtracting the first equation from the second, we get 2d = -8, leading to d = -4. Substituting the common difference back into the first equation, we find that a = 28 - d = 28 + 4 = 32.
With the first term and the common difference now known, we can attempt to maximize the sum of this sequence. Since the terms are decreasing by 4 each time, to maximize the sum of any number of terms, we should aim to include as many positive terms as possible before the terms become negative or reach zero.
The general term of the sequence is a + (n-1)d. The last positive term (before the sequence turns negative) is when a + (n-1)d > 0. Solving for n, we find that the maximum number of terms without making the sum negative is n = 9. Hence, the sum of the first 8 terms would return the maximum sum for this sequence. Using the formula for the sum of an arithmetic sequence S_n = n/2 * (2a + (n-1)d), we find the maximum sum S_8 = 4 * (64 - 28), which equals 144.