Let `a_(n)` be the nth term of an AP, if `sum_(r=1)^(100)a_(2r)= alpha " and "sum_(r=1)^(100)a_(2r-1)=beta`, then the common difference of the AP is

Options

A. (a) `alpha-beta`
B. (b) `beta-alpha`
C. (c) `(alpha-beta)/(2)`
D. (d) None of these

Explanation

To solve the problem, we need to find the common difference \( d \) of the arithmetic progression (AP) given the sums of specific terms.

### Step-by-Step Solution:

1. **Understanding the nth term of an AP**:
The nth term of an arithmetic progression can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting up the sums**:
We are given:
\[ \sum_{r=1}^{100} a_{2r} = \alpha \]
and
\[ \sum_{r=1}^{100} a_{2r-1} = \beta \]

3. **Expanding the first sum**:
The first sum can be expanded as follows:
\[ \sum_{r=1}^{100} a_{2r} = a_2 + a_4 + a_6 + \ldots + a_{200} \]
Substituting the formula for \( a_n \):
\[ = (a + d) + (a + 3d) + (a + 5d) + \ldots + (a + 199d) \]
This can be rewritten as:
\[ = 100a + (1 + 3 + 5 + \ldots + 199)d \]
The sum of the first \( n \) odd numbers is \( n^2 \). Here, \( n = 100 \), so:
\[ 1 + 3 + 5 + \ldots + 199 = 100^2 = 10000 \]
Thus, we have:
\[ \sum_{r=1}^{100} a_{2r} = 100a + 10000d = \alpha \quad \text{(Equation 1)} \]

4. **Expanding the second sum**:
The second sum can be expanded as follows:
\[ \sum_{r=1}^{100} a_{2r-1} = a_1 + a_3 + a_5 + \ldots + a_{199} \]
Substituting the formula for \( a_n \):
\[ = a + (a + 2d) + (a + 4d) + \ldots + (a + 198d) \]
This can be rewritten as:
\[ = 100a + (0 + 2 + 4 + \ldots + 198)d \]
The sum of the first \( n \) even numbers is \( n(n+1) \). Here, \( n = 100 \), so:
\[ 0 + 2 + 4 + \ldots + 198 = 100 \cdot 99 = 9900 \]
Thus, we have:
\[ \sum_{r=1}^{100} a_{2r-1} = 100a + 9900d = \beta \quad \text{(Equation 2)} \]

5. **Subtracting the two equations**:
Now, we subtract Equation 2 from Equation 1:
\[ (100a + 10000d) - (100a + 9900d) = \alpha - \beta \]
Simplifying this gives:
\[ 100d = \alpha - \beta \]

6. **Finding the common difference**:
Dividing both sides by 100, we find:
\[ d = \frac{\alpha - \beta}{100} \]

### Conclusion:
The common difference \( d \) of the AP is:
\[ d = \frac{\alpha - \beta}{100} \]