3A. The numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ are consecutive terms of an arithmetic progression, and the numbers $\cos \alpha_{1}, \cos \alpha_{2}, \ldots, \cos \alpha_{n}$ are also consecutive terms of an arithmetic progression. Determine $n$, if $\cos \alpha_{1}=\frac{1}{2}$, and $\cos \alpha_{n}=-\frac{1}{2}$.
Solution. Let $d$ be the difference of the first progression. Then:

$$
\alpha_{2}-\alpha_{1}=\alpha_{3}-\alpha_{2}=\ldots=\alpha_{n}-\alpha_{n-1}=d
$$

and for every $1<k<n$ we have $\frac{\alpha_{k-1}+\alpha_{k+1}}{2}=\alpha_{k}$. Similarly, for the second progression we have

$$
2 \cos \alpha_{k}=\cos \alpha_{k-1}+\cos \alpha_{k+1}=2 \cos \frac{\alpha_{k-1}+\alpha_{k+1}}{2} \cos \frac{\alpha_{k-1}-\alpha_{k+1}}{2}=2 \cos \alpha_{k} \cos d
$$

From this, $\cos \alpha_{k}(1-\cos d)=0$. If $1-\cos d=0$, then $d=2 m \pi, m \in \mathbb{Z}$, and from the condition $\alpha_{n}=\alpha_{1}+(n-1) d$ it follows that $\cos \alpha_{n}=\cos \alpha_{1}$, which contradicts the conditions

$$
\cos \alpha_{1}=\frac{1}{2} \text { and } \cos \alpha_{n}=-\frac{1}{2} .
$$

Therefore, $\cos \alpha_{k}=0$ for every $k, 1<k<n$, so the progression is:

$$
\cos \alpha_{1}=\frac{1}{2}, \cos \alpha_{2}=0, \cos \alpha_{3}=0, \ldots, \cos \alpha_{n-1}=0, \cos \alpha_{n}=-\frac{1}{2}
$$

and in an arithmetic progression, this is possible only if the sequence has three terms. Thus, $n=3$ and in this case the sequences are $\frac{\pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{3}$ and $\cos \frac{\pi}{3}, \cos \frac{\pi}{2}, \cos \frac{2 \pi}{3}$.