Problem (Algebra):
If $1, x, y$ is a geometric sequence and $x, y, 3$ is an arithmetic sequence, compute the maximum value of $x+y$.

Solution:
The common ratio in the geometric sequence $1, x, y$ is $\frac{x}{1}=x$, so $y=x^{2}$. The arithmetic sequence $x, y, 3$ has a common difference, so $y-x=3-y$. Substituting $y=x^{2}$ in the equation yields

$$
\begin{aligned}
x^{2}-x & =3-x^{2} \\
2 x^{2}-x-3 & =0
\end{aligned}
$$

from which $x=\frac{3}{2}$ or -1 . The respective values of $y$ are $y=x^{2}=\frac{9}{4}$ or 1 . Thus the possible values of $x+y$ are $\frac{15}{4}$ and 0 , so the answer is $\frac{\mathbf{1 5}}{\mathbf{4}}$.

Final answer: $\frac{15}{4}$