222. In the arithmetic progression $\left(a_{n}\right)$

$$
a_{k}=l, \quad a_{l}=k(k \neq l)
$$

Calculate $a_{n}$
$\triangle$ Subtract the equalities term by term

$$
l=a_{1}+d(k-1), \quad k=a_{1}+d(l-1)
$$

where $d$ is the common difference of the arithmetic progression. We get:

$$
l-k=d(k-l), \quad d=-1
$$

Then

$$
l=a_{1}-(k-1), \quad a_{1}=k+l-1
$$

It remains to find $a_{n}$ :

$$
a_{n}=a_{1}+d(n-1)=k+l-1-(n-1)=k+l-n
$$

Answer: $a_{n}=k+l-n$.