Let `a_(1), a_(2),….` and `b_(1),b_(2),….` be arithemetic progression such that `a_(1)=25`, `b_(1)=75` and `a_(100)+b_(100)=100`, then the sum of first hundred term of the progression`a_(1)+b_(1)`, `a_(2)+b_(2)`,…. is equal to

Options

A. `1000`
B. `100000`
C. `10000`
D. `24000`

Explanation

To solve the problem, we need to find the sum of the first 100 terms of the combined arithmetic progressions \(a_n\) and \(b_n\), given the initial conditions and the relationship between \(a_{100}\) and \(b_{100}\).

### Step-by-Step Solution:

1. **Identify the given values**:
- \(a_1 = 25\)
- \(b_1 = 75\)
- \(a_{100} + b_{100} = 100\)

2. **Define the general terms of the arithmetic progressions**:
- The \(n^{th}\) term of the arithmetic progression \(a_n\) can be expressed as:
\[ a_n = a_1 + (n-1)d_a \]
- The \(n^{th}\) term of the arithmetic progression \(b_n\) can be expressed as:
\[ b_n = b_1 + (n-1)d_b \]
where \(d_a\) and \(d_b\) are the common differences of the sequences \(a_n\) and \(b_n\), respectively.

3. **Calculate \(a_{100}\) and \(b_{100}\)**:
- From the definitions:
\[ a_{100} = 25 + 99d_a \]
\[ b_{100} = 75 + 99d_b \]
- Using the condition \(a_{100} + b_{100} = 100\):
\[ (25 + 99d_a) + (75 + 99d_b) = 100 \]
\[ 100 + 99(d_a + d_b) = 100 \]
\[ 99(d_a + d_b) = 0 \]
- This implies:
\[ d_a + d_b = 0 \quad \Rightarrow \quad d_b = -d_a \]

4. **Find the sum of the first 100 terms of the combined series**:
- The combined series can be expressed as:
\[ S = (a_1 + b_1) + (a_2 + b_2) + \ldots + (a_{100} + b_{100}) \]
- This can be separated into two sums:
\[ S = (a_1 + a_2 + \ldots + a_{100}) + (b_1 + b_2 + \ldots + b_{100}) \]

5. **Use the formula for the sum of an arithmetic progression**:
- The sum of the first \(n\) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (a + l) \]
- For \(a_n\):
\[ S_a = \frac{100}{2} \times (25 + a_{100}) = 50 \times (25 + (25 + 99d_a)) = 50 \times (50 + 99d_a) \]
- For \(b_n\):
\[ S_b = \frac{100}{2} \times (75 + b_{100}) = 50 \times (75 + (75 + 99d_b)) = 50 \times (150 + 99(-d_a)) \]

6. **Combine the sums**:
- Now, combine \(S_a\) and \(S_b\):
\[ S = S_a + S_b = 50 \times (50 + 99d_a) + 50 \times (150 - 99d_a) \]
\[ S = 50 \times (50 + 150) = 50 \times 200 = 10000 \]

### Final Answer:
The sum of the first 100 terms of the progression \(a_1 + b_1, a_2 + b_2, \ldots, a_{100} + b_{100}\) is **10000**.