Find the sum of even positive integers between 1 and 200.

Answer

To find the sum of even positive integers between 1 and 200, we can follow these steps:

### Step 1: Identify the even integers
The even positive integers between 1 and 200 are:
\[ 2, 4, 6, \ldots, 198 \]

### Step 2: Recognize the sequence as an Arithmetic Progression (AP)
This sequence is an arithmetic progression where:
- The first term \( a = 2 \)
- The common difference \( d = 2 \)

### Step 3: Determine the number of terms (n)
To find the number of terms \( n \), we use the formula for the \( n \)-th term of an AP:
\[ a_n = a + (n - 1) \cdot d \]
Here, \( a_n \) is the last term, which is 198. Plugging in the values:
\[ 198 = 2 + (n - 1) \cdot 2 \]

### Step 4: Solve for n
Rearranging the equation:
\[ 198 - 2 = (n - 1) \cdot 2 \]
\[ 196 = (n - 1) \cdot 2 \]
Dividing both sides by 2:
\[ 98 = n - 1 \]
Adding 1 to both sides:
\[ n = 99 \]

### Step 5: Use the sum formula for an AP
The sum \( S_n \) of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Substituting the values we have:
- \( n = 99 \)
- \( a = 2 \)
- \( d = 2 \)

### Step 6: Calculate the sum
Plugging in the values:
\[ S_{99} = \frac{99}{2} \cdot (2 \cdot 2 + (99 - 1) \cdot 2) \]
\[ S_{99} = \frac{99}{2} \cdot (4 + 98 \cdot 2) \]
\[ S_{99} = \frac{99}{2} \cdot (4 + 196) \]
\[ S_{99} = \frac{99}{2} \cdot 200 \]
\[ S_{99} = 99 \cdot 100 \]
\[ S_{99} = 9900 \]

### Final Answer
The sum of even positive integers between 1 and 200 is:
\[ \boxed{9900} \]

---