In a triangle, the measures of the angles form an arithmetic sequence. Similarly, the reciprocals of the sides also form an arithmetic sequence. Determine the angles of the triangle!
Let the angles of the triangle be denoted by $\alpha, \beta$, and $\gamma$ in the usual way, where $\alpha \leq \beta \leq \gamma$. The measures of the angles form an arithmetic sequence, and $\alpha+\beta+\gamma=180^{\circ}$, so $\beta=60^{\circ}$.

Let the foot of the perpendicular from $A$ to $BC$ be $T$. In the right triangle $ATB$, the angle at $B$ is $60^{\circ}$, thus

$$
B T=\frac{c}{2}, \quad A T=\frac{\sqrt{3}}{2} c
$$

From the right triangle $ACT$,

$$
\left(\frac{\sqrt{3}}{2} c\right)^{2}+\left(a-\frac{c}{2}\right)^{2}=b^{2}
$$

which simplifies to $a^{2}+c^{2}-a c=b^{2}$. Dividing by $ac$ and rearranging,

$$
\frac{a}{c}+\frac{c}{a}=\frac{b^{2}}{a c}+1=1+\frac{1}{4}\left(\frac{b}{a}+\frac{b}{c}\right)^{2}-\frac{1}{4}\left(\frac{b}{a}-\frac{b}{c}\right)^{2}
$$

Since $a \leq b \leq c$ and the reciprocals of the sides also form an arithmetic sequence, we have

$$
\frac{1}{b}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right)
$$

from which $\frac{b}{a}+\frac{b}{c}=2$. Substituting this into (1),

$$
\frac{a}{c}+\frac{c}{a}=2-\frac{1}{4}\left(\frac{b}{a}-\frac{b}{c}\right)^{2} \leq 2
$$

However, the sum of a positive number and its reciprocal is always at least 2, and equality holds only when the number is exactly 1. Thus, (2) can only be satisfied if $\frac{a}{c}=1$, i.e., $a=c$. But then, given $a \leq b \leq c$, all three sides of the triangle are equal. The triangle is equilateral, all angles are $60^{\circ}$, and both the angles and the reciprocals of the sides form an arithmetic sequence.