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. `(alpha-beta)/(200)`
B. `alpha-beta`
C. `(alpha-beta)/(100)`
D. `beta-alpha`

Explanation

To solve the problem, we need to find the common difference of the arithmetic progression (AP) given the sums of the even and odd indexed terms.

### Step-by-step Solution:

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

2. **Sum of the even indexed terms**:
The even indexed terms are \( a_2, a_4, a_6, \ldots, a_{200} \). We can express these terms as:
\[ a_{2r} = a + (2r-1)d \quad \text{for } r = 1, 2, \ldots, 100 \]
The sum of these terms is:
\[ \sum_{r=1}^{100} a_{2r} = \sum_{r=1}^{100} \left( a + (2r-1)d \right) = \sum_{r=1}^{100} a + \sum_{r=1}^{100} (2r-1)d \]
The first sum is:
\[ \sum_{r=1}^{100} a = 100a \]
The second sum can be simplified using the formula for the sum of the first n natural numbers:
\[ \sum_{r=1}^{100} (2r-1) = 2\sum_{r=1}^{100} r - \sum_{r=1}^{100} 1 = 2 \cdot \frac{100 \cdot 101}{2} - 100 = 10100 - 100 = 10000 \]
Thus, we have:
\[ \sum_{r=1}^{100} a_{2r} = 100a + 10000d = \alpha \]

3. **Sum of the odd indexed terms**:
The odd indexed terms are \( a_1, a_3, a_5, \ldots, a_{199} \). We can express these terms as:
\[ a_{2r-1} = a + (2r-2)d \quad \text{for } r = 1, 2, \ldots, 100 \]
The sum of these terms is:
\[ \sum_{r=1}^{100} a_{2r-1} = \sum_{r=1}^{100} \left( a + (2r-2)d \right) = \sum_{r=1}^{100} a + \sum_{r=1}^{100} (2r-2)d \]
The first sum is:
\[ \sum_{r=1}^{100} a = 100a \]
The second sum can be simplified as:
\[ \sum_{r=1}^{100} (2r-2) = 2\sum_{r=1}^{100} r - 2\sum_{r=1}^{100} 1 = 2 \cdot \frac{100 \cdot 101}{2} - 200 = 10100 - 200 = 9900 \]
Thus, we have:
\[ \sum_{r=1}^{100} a_{2r-1} = 100a + 9900d = \beta \]

4. **Setting up the equations**:
We now have two equations:
\[ 100a + 10000d = \alpha \quad \text{(1)} \]
\[ 100a + 9900d = \beta \quad \text{(2)} \]

5. **Subtracting the equations**:
Subtract equation (2) from equation (1):
\[ (100a + 10000d) - (100a + 9900d) = \alpha - \beta \]
This simplifies to:
\[ 100d = \alpha - \beta \]
Therefore, we can solve for \( d \):
\[ d = \frac{\alpha - \beta}{100} \]

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