The sum of $m$ terms of an arithmetic progression is $n$ and the sum of $n$ terms is $m$. Calculate the sum of $m+n$ and $m-n$ terms of this progression.

Let's denote the first term of the arithmetic progression by $a$ and the common difference by $d$. The sum of the first $k$ terms of an arithmetic progression is given by the formula:
\[ S_k = \frac{k}{2} \left(2a + (k-1)d\right) \]

Given:
\[ S_m = \frac{m}{2} \left(2a + (m-1)d\right) = n \]
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) = m \]

We need to find the sum of $m+n$ and $m-n$ terms:
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]
\[ S_{m-n} = \frac{m-n}{2} \left(2a + (m-n-1)d\right) \]

First, let's solve the given equations for $a$ and $d$.

From the first equation:
\[ \frac{m}{2} \left(2a + (m-1)d\right) = n \]
\[ 2a + (m-1)d = \frac{2n}{m} \]
\[ 2a + (m-1)d = \frac{2n}{m} \quad \text{(1)} \]

From the second equation:
\[ \frac{n}{2} \left(2a + (n-1)d\right) = m \]
\[ 2a + (n-1)d = \frac{2m}{n} \]
\[ 2a + (n-1)d = \frac{2m}{n} \quad \text{(2)} \]

Subtract equation (2) from equation (1):
\[ (2a + (m-1)d) - (2a + (n-1)d) = \frac{2n}{m} - \frac{2m}{n} \]
\[ (m-1)d - (n-1)d = \frac{2n^2 - 2m^2}{mn} \]
\[ (m-n)d = \frac{2(n^2 - m^2)}{mn} \]
\[ (m-n)d = \frac{2(n-m)(n+m)}{mn} \]
\[ d = \frac{2(n+m)}{mn} \]

Now substitute \( d \) back into equation (1):
\[ 2a + (m-1) \left( \frac{2(n+m)}{mn} \right) = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(n+m)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(mn + m^2 - n - m)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2mn + 2m^2 - 2n - 2m}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2m^2 - 2m + 2mn - 2n}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2m(m-1) + 2n(m-1)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]
\[ 2a + \frac{2(m-1)(m+n)}{mn} = \frac{2n}{m} \]

Now, we can find \( S_{m+n} \) and \( S_{m-n} \):

\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1) \frac{2(n+m)}{mn} \right) \]
\[ S_{m+n} = \frac{m+n}{2} \left(2a + \frac{2(m+n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} = \frac{m+n}{2} \left(2a + \frac{2(m+n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} = \frac{m+n}{2} \left(2a + \frac{2(m+n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} = \frac{m+n}{2} \left(2a + \frac{2(m+n-1)(m+n)}{mn} \right) \]

Similarly, for \( S_{m-n} \):

\[ S_{m-n} = \frac{m-n}{2} \left(2a + (m-n-1)d\right) \]
\[ S_{m-n} = \frac{m-n}{2} \left(2a + (m-n-1) \frac{2(n+m)}{mn} \right) \]
\[ S_{m-n} = \frac{m-n}{2} \left(2a + \frac{2(m-n-1)(m+n)}{mn} \right) \]
\[ S_{m-n} = \frac{m-n}{2} \left(2a + \frac{2(m-n-1)(m+n)}{mn} \right) \]

Finally, the sum of \( S_{m+n} \) and \( S_{m-n} \) is:

\[ S_{m+n} + S_{m-n} = \frac{m+n}{2} \left(2a + \frac{2(m+n-1)(m+n)}{mn} \right) + \frac{m-n}{2} \left(2a + \frac{2(m-n-1)(m+n)}{mn} \right) \]

Simplifying, we get:

\[ S_{m+n} + S_{m-n} = \frac{m+n + m-n}{2} \left(2a + \frac{2(m+n-1)(m+n) + 2(m-n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = \frac{2m}{2} \left(2a + \frac{2(m+n-1)(m+n) + 2(m-n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{2(m+n-1)(m+n) + 2(m-n-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{2(m+n)(m+n-1 + m-n-1)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{2(m+n)(2m-2)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{4(m+n)(m-1)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{4(m-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{4(m-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{4(m-1)(m+n)}{mn} \right) \]
\[ S_{m+n} + S_{m-n} = m \left(2a + \frac{4(m-1)(m+n)}{mn} \right) \]

Since \( 2a + \frac{4(m-1)(m+n)}{mn} = 0 \):

\[ S_{m+n} + S_{m-n} = 0 \]

Thus, the sum of \( m+n \) and \( m-n \) terms of this arithmetic progression is:

\[ \boxed{0} \]
Based on the well-known sum formula:

$$
S_{m}=n=m a+m(m-1) \frac{d}{2}, S_{n}=m=n a+n(n-1) \frac{d}{2}
$$

from which

$$
d=-2 \frac{m+n}{m n}
$$

Similarly,

$$
\begin{gathered}
S_{m+n}=(m+n) a+(m+n)(m+n-1) \frac{d}{2}= \\
=m a+m(m-1) \frac{d}{2}+n a+n(n-1) \frac{d}{2}+m n d= \\
=S_{m}+S_{n}+m n d=m+n-2(m+n)= \\
=-(m+n) \\
S_{m-n}=(m-n) a+(m-n)(m-n-1) \frac{d}{2}= \\
=m a+m(m-1) \frac{d}{2}-n a-n(n-1) \frac{d}{2}-m n d+2 n^{2} \frac{d}{2}= \\
=S_{m}-S_{n}-d n(m-n)= \\
=\frac{(m-n)(2 n+m)}{m} .
\end{gathered}
$$

The problem was solved by: Bayer B., Burján K., Czank K., Faith F., Feldmár V., Filkorn J., Izsáky L., Kerekes T., Kertész G., König D, Krausz B, Krisztián Gy., Lukhaub Gy., Perl Gy., Pilczer P., Póka Gy., Rosenberg Á., Scharff J., Scheuer R., Smodics K., Szmodics H., Tézner O., Weisz A., Wohlstein S.