If `S_(1) = a_(2) + a_(4) + a_(6)+…` up to 100 terms and `S_(2) = a_(1) + a_(3) + a_(5) +….` upto 100 terms of a certain A.P. then its common difference d is

Options

A. `S_(1) - S_(2)`
B. `S_(2) - S_(1)`
C. `(S_(1) - S_(2))/(2)`
D. none of these

Answer

To solve the problem, we need to find the common difference \( d \) of the arithmetic progression (A.P.) given the sums \( S_1 \) and \( S_2 \).

### Step-by-Step Solution:

1. **Understand the Terms**:
- In an A.P., the \( n \)-th term is given by:
\[ a_n = a + (n-1)d \]
- Here, \( a \) is the first term and \( d \) is the common difference.

2. **Calculate \( S_1 \)**:
- \( S_1 \) is the sum of the even-indexed terms up to 100 terms:
\[ S_1 = a_2 + a_4 + a_6 + \ldots + a_{200} \]
- The even-indexed terms can be expressed as:
\[ S_1 = (a + d) + (a + 3d) + (a + 5d) + \ldots + (a + 199d) \]
- This is an A.P. with:
- First term \( = a + d \)
- Last term \( = a + 199d \)
- Number of terms \( = 100 \)
- The sum of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (\text{first term} + \text{last term}) \]
- Thus,
\[ S_1 = \frac{100}{2} \times \left((a + d) + (a + 199d)\right) = 50 \times (2a + 200d) = 100a + 100d \]

3. **Calculate \( S_2 \)**:
- \( S_2 \) is the sum of the odd-indexed terms up to 100 terms:
\[ S_2 = a_1 + a_3 + a_5 + \ldots + a_{199} \]
- The odd-indexed terms can be expressed as:
\[ S_2 = a + (a + 2d) + (a + 4d) + \ldots + (a + 198d) \]
- This is also an A.P. with:
- First term \( = a \)
- Last term \( = a + 198d \)
- Number of terms \( = 100 \)
- Therefore,
\[ S_2 = \frac{100}{2} \times (a + (a + 198d)) = 50 \times (2a + 198d) = 100a + 99d \]

4. **Find \( S_1 - S_2 \)**:
- Now, we subtract \( S_2 \) from \( S_1 \):
\[ S_1 - S_2 = (100a + 100d) - (100a + 99d) = 100d - 99d = d \]

5. **Conclusion**:
- Therefore, the common difference \( d \) can be expressed as:
\[ d = S_1 - S_2 \]

### Final Answer:
The common difference \( d \) is given by:
\[ d = S_1 - S_2 \]