The maximum sum of the series ` 100 + 98 + 96 + …` is

Options

A. `2500
B. 2550
C. 2050
D. 2555

Answer

To find the maximum sum of the series \(100 + 98 + 96 + \ldots\), we can follow these steps:

### Step 1: Identify the series
The series is an arithmetic series where the first term \(a = 100\) and the common difference \(d = 98 - 100 = -2\).

### Step 2: Determine the last term
The series continues until it reaches the smallest non-negative integer. The last term can be found by setting the general term of the arithmetic series equal to zero:
\[ a_n = a + (n-1)d \]
Setting \(a_n = 0\):
\[ 0 = 100 + (n-1)(-2) \]
\[ 0 = 100 - 2(n-1) \]
\[ 2(n-1) = 100 \]
\[ n-1 = 50 \implies n = 51 \]
Thus, there are 51 terms in the series.

### Step 3: Use the formula for the sum of an arithmetic series
The formula for the sum \(S_n\) of the first \(n\) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Substituting the values we found:
- \(n = 51\)
- \(a = 100\)
- \(d = -2\)

### Step 4: Calculate the sum
Substituting into the formula:
\[ S_{51} = \frac{51}{2} \times (2 \times 100 + (51 - 1)(-2)) \]
\[ = \frac{51}{2} \times (200 - 100) \]
\[ = \frac{51}{2} \times 100 \]
\[ = 51 \times 50 = 2550 \]

### Conclusion
The maximum sum of the series \(100 + 98 + 96 + \ldots\) is \(2550\).

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