To find the sum of the arithmetic series \( A = 2 + 4 + 6 + \ldots + 98 \), we can use three different methods. Let's go through each method step-by-step.

### Method 1: Using the Formula for the Sum of an Arithmetic Series

The sum \( S_n \) of the first \( n \) terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (a + l) \]
where:
- \( n \) is the number of terms,
- \( a \) is the first term,
- \( l \) is the last term.

For the series \( 2 + 4 + 6 + \ldots + 98 \):
- The first term \( a = 2 \).
- The last term \( l = 98 \).
- The common difference \( d = 2 \).

First, we need to find the number of terms \( n \). The \( n \)-th term of an arithmetic series is given by:
\[ a_n = a + (n-1)d \]

Setting \( a_n = 98 \):
\[ 98 = 2 + (n-1) \cdot 2 \]
\[ 98 = 2 + 2n - 2 \]
\[ 98 = 2n \]
\[ n = 49 \]

Now, we can use the sum formula:
\[ S_{49} = \frac{49}{2} \times (2 + 98) \]
\[ S_{49} = \frac{49}{2} \times 100 \]
\[ S_{49} = 49 \times 50 \]
\[ S_{49} = 2450 \]

### Method 2: Pairing Terms

We can pair the terms from the beginning and the end of the series to simplify the calculation. Each pair will have the same sum.

Pairs:
\[ (2 + 98), (4 + 96), (6 + 94), \ldots, (48 + 52), (50) \]

Each pair sums to 100, and there are \( \frac{49}{2} \) pairs (since there are 49 terms, one term will be unpaired, which is 50).

\[ \text{Number of pairs} = 24 \text{ pairs} + 1 \text{ unpaired term} \]

Sum of pairs:
\[ 24 \times 100 + 50 = 2400 + 50 = 2450 \]

### Method 3: Using the Formula for the Sum of the First \( n \) Even Numbers

The sum of the first \( n \) even numbers is given by:
\[ S_n = n(n + 1) \]

For the series \( 2 + 4 + 6 + \ldots + 98 \), we have already determined that there are 49 terms.

\[ S_{49} = 49 \times (49 + 1) \]
\[ S_{49} = 49 \times 50 \]
\[ S_{49} = 2450 \]

### Conclusion

Using all three methods, we find that the sum of the arithmetic series \( 2 + 4 + 6 + \ldots + 98 \) is:
\[ \boxed{2450} \]