Let's determine the first term and the common difference of an arithmetic progression, if the sum of the first $n$ terms of this progression is $\frac{n^{2}}{2}$ for all values of $n$.
I. Solution. Since the sum of the first $n$ terms is $\frac{n^{2}}{2}$, the first term is $\frac{1^{2}}{2}=\frac{1}{2}$; the sum of the first two terms is $\frac{2^{2}}{2}=2$; thus, the second term is $2-\frac{1}{2}=\frac{3}{2}$, and the common difference $d=1$; the arithmetic sequence itself is:

$$
\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots, \frac{2 n-1}{2}
$$

(Dénes König, Budapest.)

II. Solution. According to the problem:

$$
\frac{2 a_{1}+(n-1) d}{2} \cdot n=\frac{n^{2}}{2}
$$

from which

$$
n(d-1)+2 a_{1}-d=0
$$

for the expression on the left to be 0 for all values of $n$, it is necessary that

$$
d-1=0 \quad \text{and} \quad 2 a_{1}-d=0
$$

be true. From the first equation, $d=1$, and from the second, $a_{1}=\frac{1}{2}$.

(Sándor Jankovich, Budapest.)

The problem was also solved by: Appel S., Bayer B., Bender E., Benedek Zs., Czank K., Faith F., Filkorn J., Freibauer E., Glass M., Groffits A., Kerekes T., Kohn B., Krausz B., Krausz J., Gy. Krisztián, Gy. Lukhaub, R. Obláth, F. Pálfy, Gy. Perl, J. Porkoláb, P. Rozlosnik, G. Sasvári, J. Sasvári, Ö. Spiczer, F. Stromfeld, J. Weisz.