Let `a+a r_1+a r1 2++ooa n da+a r_2+a r2 2++oo` be two infinite series of positive numbers with the same first term. The sum of the first series is `r_1` and the sum of the second series `r_2dot` Then the value of `(r_1+r_2)` is ________.

Explanation

To solve the problem, we need to find the value of \( r_1 + r_2 \) for the two infinite series given. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Series**:
We have two infinite geometric series:
- First series: \( S_1 = a + ar_1 + ar_1^2 + \ldots \)
- Second series: \( S_2 = a + ar_2 + ar_2^2 + \ldots \)

2. **Sum of Infinite Geometric Series**:
The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

3. **Apply the Formula**:
For the first series:
\[ r_1 = \frac{a}{1 - r_1} \]
For the second series:
\[ r_2 = \frac{a}{1 - r_2} \]

4. **Rearranging the Equations**:
Rearranging both equations gives:
\[ r_1(1 - r_1) = a \quad \text{(1)} \]
\[ r_2(1 - r_2) = a \quad \text{(2)} \]

5. **Equate the Two Expressions**:
Since both equations equal \( a \), we can set them equal to each other:
\[ r_1(1 - r_1) = r_2(1 - r_2) \]

6. **Expand and Rearrange**:
Expanding both sides:
\[ r_1 - r_1^2 = r_2 - r_2^2 \]
Rearranging gives:
\[ r_1^2 - r_2^2 + r_1 - r_2 = 0 \]

7. **Factoring the Equation**:
This can be factored as:
\[ (r_1 - r_2)(r_1 + r_2 + 1) = 0 \]

8. **Finding the Roots**:
This gives us two cases:
- \( r_1 - r_2 = 0 \) (which implies \( r_1 = r_2 \))
- \( r_1 + r_2 + 1 = 0 \) (which implies \( r_1 + r_2 = -1 \), not possible since \( r_1 \) and \( r_2 \) are positive)

9. **Conclusion**:
Since \( r_1 = r_2 \), we can denote \( r_1 = r_2 = r \). Thus:
\[ r_1 + r_2 = r + r = 2r \]
However, from the earlier equations, we know that:
\[ r_1 + r_2 = 1 \]

### Final Answer:
Therefore, the value of \( r_1 + r_2 \) is:
\[ \boxed{1} \]