Let t, denote the rth term in the binomial expansion of `(1 + a)^(50)`. If
`t_(25)+t_(27)=125/52t_(26)`
then the sum of all possible values of a is ___

Answer

To solve the problem, we need to analyze the given equation involving the rth term in the binomial expansion of \((1 + a)^{50}\).

### Step-by-Step Solution:

1. **Identify the rth term formula**:
The rth term \( t_r \) in the binomial expansion of \((1 + a)^{50}\) is given by:
\[ t_r = \binom{50}{r-1} \cdot 1^{50 - (r-1)} \cdot a^{r-1} = \binom{50}{r-1} a^{r-1} \]

2. **Express the terms \( t_{25}, t_{26}, t_{27} \)**:
- For \( t_{25} \):
\[ t_{25} = \binom{50}{24} a^{24} \]
- For \( t_{26} \):
\[ t_{26} = \binom{50}{25} a^{25} \]
- For \( t_{27} \):
\[ t_{27} = \binom{50}{26} a^{26} \]

3. **Set up the equation**:
According to the problem, we have:
\[ t_{25} + t_{27} = \frac{125}{52} t_{26} \]
Substituting the expressions for \( t_{25}, t_{26}, t_{27} \):
\[ \binom{50}{24} a^{24} + \binom{50}{26} a^{26} = \frac{125}{52} \binom{50}{25} a^{25} \]

4. **Factor out common terms**:
We can factor out \( a^{24} \) from the left-hand side:
\[ a^{24} \left( \binom{50}{24} + \binom{50}{26} a^2 \right) = \frac{125}{52} \binom{50}{25} a^{25} \]

5. **Rearranging the equation**:
Dividing both sides by \( a^{24} \) (assuming \( a \neq 0 \)):
\[ \binom{50}{24} + \binom{50}{26} a^2 = \frac{125}{52} \binom{50}{25} a \]

6. **Rearranging into a standard quadratic form**:
Rearranging gives us:
\[ \binom{50}{26} a^2 - \frac{125}{52} \binom{50}{25} a + \binom{50}{24} = 0 \]

7. **Identify coefficients**:
Let:
- \( A = \binom{50}{26} \)
- \( B = -\frac{125}{52} \binom{50}{25} \)
- \( C = \binom{50}{24} \)

8. **Use the quadratic formula**:
The roots of the quadratic equation \( Ax^2 + Bx + C = 0 \) can be found using:
\[ a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

9. **Calculate the discriminant**:
We need to ensure that the discriminant \( B^2 - 4AC \) is non-negative for real roots.

10. **Sum of the roots**:
The sum of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \text{Sum} = -\frac{B}{A} \]

11. **Substituting back**:
Substitute \( B \) and \( A \) into the sum formula to find the sum of all possible values of \( a \).

12. **Final calculation**:
After calculating, we find the sum of all possible values of \( a \) is:
\[ 2 + \frac{1}{2} = 2.5 \]

### Final Answer:
The sum of all possible values of \( a \) is \( \boxed{2.5} \).