Three numbers have sum $k$ (where $k\in \mathbb{R}$) such that the numbers are arethmetic progression.If First of two numbers remain the same and to the third number we add $\frac{k}{6}$ than we have geometry progression.
Find those numbers?
1. Let the three numbers in arithmetic progression be \(a - d\), \(a\), and \(a + d\). Given that their sum is \(k\), we have:
   \[
   (a - d) + a + (a + d) = k
   \]
   Simplifying, we get:
   \[
   3a = k \implies a = \frac{k}{3}
   \]
   Therefore, the three numbers are:
   \[
   \frac{k}{3} - d, \frac{k}{3}, \frac{k}{3} + d
   \]

2. According to the problem, if we keep the first two numbers the same and add \(\frac{k}{6}\) to the third number, the sequence becomes a geometric progression. Thus, the new sequence is:
   \[
   \frac{k}{3} - d, \frac{k}{3}, \frac{k}{3} + d + \frac{k}{6}
   \]
   Simplifying the third term:
   \[
   \frac{k}{3} + d + \frac{k}{6} = \frac{k}{3} + \frac{k}{6} + d = \frac{k}{2} + d
   \]

3. For these numbers to be in geometric progression, the square of the middle term must equal the product of the first and third terms:
   \[
   \left(\frac{k}{3}\right)^2 = \left(\frac{k}{3} - d\right) \left(\frac{k}{2} + d\right)
   \]
   Expanding the right-hand side:
   \[
   \frac{k^2}{9} = \left(\frac{k}{3} - d\right) \left(\frac{k}{2} + d\right)
   \]
   \[
   \frac{k^2}{9} = \frac{k}{3} \cdot \frac{k}{2} + \frac{k}{3} \cdot d - d \cdot \frac{k}{2} - d^2
   \]
   \[
   \frac{k^2}{9} = \frac{k^2}{6} + \frac{k}{6}d - d^2
   \]

4. Rearranging the equation to form a quadratic in \(d\):
   \[
   \frac{k^2}{9} = \frac{k^2}{6} + \frac{k}{6}d - d^2
   \]
   \[
   d^2 + \frac{k}{6}d - \frac{k^2}{18} = 0
   \]

5. Solving this quadratic equation using the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = \frac{k}{6}\), and \(c = -\frac{k^2}{18}\):
   \[
   d = \frac{-\frac{k}{6} \pm \sqrt{\left(\frac{k}{6}\right)^2 - 4 \cdot 1 \cdot \left(-\frac{k^2}{18}\right)}}{2 \cdot 1}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{4k^2}{18}}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{2k^2}{9}}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \sqrt{\frac{k^2}{36} + \frac{8k^2}{36}}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \sqrt{\frac{9k^2}{36}}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \frac{3k}{6}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} \pm \frac{k}{2}}{2}
   \]
   \[
   d = \frac{-\frac{k}{6} + \frac{k}{2}}{2} \quad \text{or} \quad d = \frac{-\frac{k}{6} - \frac{k}{2}}{2}
   \]
   \[
   d = \frac{\frac{k}{3}}{2} = \frac{k}{6} \quad \text{or} \quad d = \frac{-\frac{2k}{3}}{2} = -\frac{k}{3}
   \]

6. Therefore, the possible values for \(d\) are \(\frac{k}{6}\) and \(-\frac{k}{3}\). Substituting these back into the original arithmetic progression:
   - For \(d = \frac{k}{6}\):
     \[
     \frac{k}{3} - \frac{k}{6}, \frac{k}{3}, \frac{k}{3} + \frac{k}{6} \implies \frac{k}{6}, \frac{k}{3}, \frac{k}{2}
     \]
   - For \(d = -\frac{k}{3}\):
     \[
     \frac{k}{3} - \left(-\frac{k}{3}\right), \frac{k}{3}, \frac{k}{3} + \left(-\frac{k}{3}\right) \implies \frac{2k}{3}, \frac{k}{3}, 0
     \]

The final answer is \(\boxed{\left(\frac{k}{6}, \frac{k}{3}, \frac{k}{2}\right)}\) or \(\boxed{\left(\frac{2k}{3}, \frac{k}{3}, 0\right)}\).