## Task B-4.3.

Three different real numbers $a$, 2016, and $b$ are three consecutive terms of a geometric sequence. If the numbers $a+2016, b+2016$, and $a+b$ are three consecutive terms of an arithmetic sequence, determine the numbers $a$ and $b$.
## Solution.

$a+2016, b+2016, a+b$ is an arithmetic sequence, so

$$
\begin{aligned}
b+2016 & =\frac{a+2016+a+b}{2} \\
a & =\frac{b+2016}{2}
\end{aligned}
$$

$a, 2016, b$ is a geometric sequence with $q=\frac{2016}{a}=\frac{b}{2016}$.

It follows that $a b=2016^{2}$.

$$
\begin{aligned}
& \frac{b+2016}{2} \cdot b=2016^{2} \\
& b^{2}+2016 b-2 \cdot 2016^{2}=0
\end{aligned}
$$

It follows that

$$
\begin{array}{cl}
b=\frac{-2016 \pm \sqrt{2016^{2}+8 \cdot 2016^{2}}}{2}=\frac{-2016 \pm 3 \cdot 2016}{2} & 1 \text{ point} \\
b_{1}=2016, b_{2}=-4032 . & 1 \text{ point}
\end{array}
$$

From $b \neq 2016$ it follows that $b=-4032$. Then $a=\frac{b+2016}{2}=-1008$.