1. [5] Determine all pairs $(a, b)$ of real numbers such that $10, a, b, a b$ is an arithmetic progression.
Answer: $(4,-2),\left(\frac{5}{2},-5\right)$ Since $10, a, b$ is an arithmetic progression, we have $a=\frac{1}{2}(10+b)$. Also, we have $a+a b=2 b$, and so $a(1+b)=2 b$. Substituting the expression for $a$ gives $(10+b)(1+b)=4 b$. Solving this quadratic equation gives the solutions $b=-2$ and $b=-5$. The corresponding values for $a$ can be found by $a=\frac{1}{2}(10+b)$.