What will be the output of the following code?

1 numbers {1,2,3,4,5,5,6}

2 int num1 numbers.size() 1;

3 int num2 0;

4 for(each number){

5 add number to num2 and assign again to num2

6}

7 int num3 = num2 (num1* (num1+1)/2);

8 return num3;

Let's break down the code step by step to understand what it does and calculate the output.

### Step-by-Step Execution:

1. **Initialization**:
   - `numbers = {1, 2, 3, 4, 5, 5, 6}`
   - `num1 = numbers.size() - 1`
     - `numbers.size()` returns the size of the list, which is 7.
     - Therefore, `num1 = 7 - 1 = 6`.
   - `num2 = 0`

2. **For Loop**:
   - The loop iterates over each element in `numbers`, adding each element to `num2`.
   - Let's calculate the sum:

   \[
   \text{num2} = 1 + 2 + 3 + 4 + 5 + 5 + 6 = 26
   \]

3. **Calculation of `num3`**:
   - The formula used here is:

   \[
   \text{num3} = \text{num2} - \left(\frac{\text{num1} \times (\text{num1} + 1)}{2}\right)
   \]

   - First, calculate the value inside the parentheses:

   \[
   \text{num1} \times (\text{num1} + 1) = 6 \times 7 = 42
   \]

   - Then, divide by 2:

   \[
   \frac{42}{2} = 21
   \]

   - Finally, subtract this from `num2`:

   \[
   \text{num3} = 26 - 21 = 5
   \]

### Output:
The final output of the code will be **5**.

**Would you like more details or have any questions?**

Here are some related questions to consider:
1. What is the purpose of the formula used to calculate `num3`?
2. What would the output be if there were no duplicate values in `numbers`?
3. How does changing the order of elements in `numbers` affect the output?
4. How can you modify the code to find the sum of unique elements only?
5. What will happen if the list `numbers` is empty?

**Tip**: The formula used in the code is often related to detecting duplicates in a sequence by comparing the sum with the expected sum of a sequence without duplicates.

Explore the solution to calculating the output of an arithmetic series code example step-by-step. Learn about summation and arithmetic operations. Suitable for middle school students.

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