Let `a_(1), a_(2), a_(3).... and b_(1), b_(2), b_(3)...` be arithmetic progression such that `a_(1) = 25, b_(1) = 75 and a_(100) + b_(100) = 100`. Then

Options

A. the difference between successive terms in progression 'a' is opposite of the difference in progression 'b'
B. `a_(n) + b_(n) = 100` for any n
C. `(a_(1) + b_(1)), (a_(2) + b_(2)), (a_(3) + b_(3)),`..... are in A.P.
D. `underset(r = 1)overset(100)sum (a_(r) + b_(r)) = 10000`

Answer

To solve the problem, we need to analyze the arithmetic progressions given the conditions provided.

### Step 1: Define the arithmetic progressions
Let the first arithmetic progression be \( a_n \) with:
- First term \( a_1 = 25 \)
- Common difference \( d_A \)

Let the second arithmetic progression be \( b_n \) with:
- First term \( b_1 = 75 \)
- Common difference \( d_B \)

### Step 2: Write the general terms for \( a_n \) and \( b_n \)
The general term for the first arithmetic progression \( a_n \) can be expressed as:
\[ a_n = a_1 + (n-1) d_A = 25 + (n-1) d_A \]

The general term for the second arithmetic progression \( b_n \) can be expressed as:
\[ b_n = b_1 + (n-1) d_B = 75 + (n-1) d_B \]

### Step 3: Find \( a_{100} \) and \( b_{100} \)
Using the formulas derived:
\[ a_{100} = 25 + (100 - 1) d_A = 25 + 99 d_A \]
\[ b_{100} = 75 + (100 - 1) d_B = 75 + 99 d_B \]

### Step 4: Set up the equation based on the given condition
According to the problem, we have:
\[ a_{100} + b_{100} = 100 \]
Substituting the expressions we found:
\[ (25 + 99 d_A) + (75 + 99 d_B) = 100 \]
Simplifying this gives:
\[ 100 + 99 d_A + 99 d_B = 100 \]
Subtracting 100 from both sides:
\[ 99 d_A + 99 d_B = 0 \]
Dividing through by 99:
\[ d_A + d_B = 0 \]
This implies:
\[ d_A = -d_B \]

### Step 5: Analyze \( a_n + b_n \)
Now, we want to find \( a_n + b_n \):
\[ a_n + b_n = (25 + (n-1) d_A) + (75 + (n-1) d_B) \]
Combining the terms:
\[ a_n + b_n = 100 + (n-1)(d_A + d_B) \]
Since we found that \( d_A + d_B = 0 \):
\[ a_n + b_n = 100 + (n-1) \cdot 0 = 100 \]

### Conclusion
Thus, for any \( n \), \( a_n + b_n = 100 \).

### Step 6: Find the sum from \( r = 1 \) to \( 100 \)
We need to find the sum:
\[ \sum_{r=1}^{100} (a_r + b_r) \]
Since \( a_r + b_r = 100 \) for all \( r \):
\[ \sum_{r=1}^{100} (a_r + b_r) = 100 + 100 + 100 + \ldots \text{ (100 times)} = 100 \times 100 = 10,000 \]

### Final Answer
The final answer is:
\[ \boxed{10000} \]