5. The sequence of numbers $a_{1}, a_{2}, \ldots, a_{2016}, \quad$ is a geometric progression, and the sequence $b_{1}, b_{2}, \ldots, b_{2016}$ is an arithmetic progression. It is known that among all the quadratic polynomials $P_{i}(x)=x^{2}+a_{i} x+b_{i}, i=1, \ldots, 2016$ only one quadratic polynomial $\mathrm{P}_{k}(x)$ has real roots. Find all possible values of $k$.
Solution: If $a_{1}=3, a_{2}=1, \ldots, a_{2016}=\frac{1}{3^{2016}}, b_{1}=1, b_{2}=2, \ldots, b_{2016}=2016$, then $k=1$. If $a_{1}=\frac{1}{3^{2016}}, \quad a_{2}=\frac{1}{3^{2015}}, \ldots, \quad a_{2015}=1, \quad a_{2016}=3, \quad b_{1}=2016, b_{2}=2015, \ldots, \quad b_{2015}=2, \quad b_{2016}=1$, then $k=2016$.

Now let $1<k<2016$. Then the quadratic polynomial $\mathrm{P}_{k}(x)$ has real roots (i.e., $\mathrm{D}_{k}=a_{k}^{2}-4 b_{k} \geq 0$), while the others do not. Consider the quadratic polynomials $\mathrm{P}_{k-1}(x)$ and $\mathrm{P}_{k+1}(x)$.

$\mathrm{D}_{k-1}=a_{k-1}^{2}-4 b_{k-1}<0$ and $\mathrm{D}_{k+1}=a_{k+1}^{2}-4 b_{k+1}<0$. Then on one hand

$\mathrm{D}_{k-1}+\mathrm{D}_{k+1}<0$, and on the other hand $\mathrm{D}_{k-1}+\mathrm{D}_{k+1}=a_{k-1}^{2}+a_{k+1}^{2}-4\left(b_{k-1}+b_{k+1}\right) \geq 2 a_{k-1} a_{k+1}-4\left(b_{k-1}+b_{k+1}\right)$ (using $a_{k-1}^{2}+a_{k+1}^{2} \geq 2 a_{k-1} a_{k+1}$). Since $a_{1}, a_{2}, \ldots, a_{2016}$ is a geometric progression, then $a_{k}^{2}=a_{k-1} a_{k+1}$, and the sequence $b_{1}, b_{2}, \ldots, b_{2016}$ is an arithmetic progression, then $b_{k}=\frac{b_{k-1}+b_{k+1}}{2}$. We get that $\quad \mathrm{D}_{k-1}+\mathrm{D}_{k+1} \geq 2 a_{k}^{2}-8 b_{k}=2 D_{k} \geq 0$, which is a contradiction. Therefore, $k$ can only take the values 1 and 2016.

## Answer: $1 ; 2016$

## Recommendations for checking.

If the correct solution provides examples of sequences for values $k=1$ and $k=2016$ - 5 points.

If only the values $k=1$ and $k=2016$ are given without examples of sequences - 0 points.

If only the values $k=1$ and $k=2016$ are given with examples of sequences - 2 points.