`-4, 0, 2, 3`
A sequence of numbers is formed by repeating the set of numbers until 80 numbers have been listed. What is the sum of the first 31 terms of the sequence?

Answer

To find the sum of the first 31 terms of the sequence formed by repeating the numbers \(-4, 0, 2, 3\), we can follow these steps:

### Step 1: Identify the repeating pattern
The sequence is formed by repeating the numbers \(-4, 0, 2, 3\). This means every four terms, the same set of numbers appears again.

### Step 2: Determine how many complete groups of 4 fit into 31 terms
Since the sequence repeats every 4 terms, we can calculate how many complete groups of 4 fit into 31 terms:
\[ 31 \div 4 = 7 \quad \text{(complete groups)} \quad \text{with a remainder of } 3. \]
This means we have 7 complete groups of 4 terms, plus 3 additional terms.

### Step 3: Calculate the sum of one complete group
Next, we calculate the sum of one complete group of the numbers:
\[ \text{Sum of one group} = -4 + 0 + 2 + 3 = 1. \]

### Step 4: Calculate the sum of the complete groups
Since we have 7 complete groups, the total sum from these groups is:
\[ \text{Sum from 7 groups} = 7 \times 1 = 7. \]

### Step 5: Add the sum of the remaining terms
Now, we need to add the sum of the first 3 terms from the next group. The first 3 terms of the sequence are:
\[ -4, 0, 2. \]
Calculating their sum:
\[ \text{Sum of remaining terms} = -4 + 0 + 2 = -2. \]

### Step 6: Calculate the total sum
Finally, we combine the sums from the complete groups and the remaining terms:
\[ \text{Total sum} = 7 + (-2) = 5. \]

Thus, the sum of the first 31 terms of the sequence is:
\[ \boxed{5}. \]