6.2 Find the number of terms of an arithmetic progression, the sum of all of which is 112, the product of the second term by the common difference is 30, and the sum of the third and fifth terms is 32. Write the first three terms of this progression.
6.2 Using the condition and formula (6.1), we have the system

$$
\left\{\begin{array} { l } 
{ a _ { 2 } d = 3 0 , } \\
{ a _ { 3 } + a _ { 5 } = 3 2 , }
\end{array} \text { or } \left\{\begin{array}{l}
\left(a_{1}+d\right) d=30 \\
a_{1}+2 d+a_{1}+4 d=32
\end{array}\right.\right.
$$

From here, we find two solutions: 1) $a_{1}=1, d=5 ; 2$ ) $a_{1}=7, d=3$. For each of them, we will use formula (6.3).

1) When $a_{1}=1, d=5$, we get $112=\frac{2+5(n-1)}{2} n$, from which $n_{1}=7, n_{2}=-6.4$ (not suitable). In this case, the first three terms are: $1 ; 6 ; 11$.
2) When $a_{1}=7, d=3$, we get $112=\frac{14+3(n-1)}{2} n$, from which $n_{1}=7, n_{2}=-\frac{32}{3}$ (not suitable). In this case, the first three terms are: $7 ; 10 ; 13$. Note that in both cases, the number of terms is 7.

Answer: 7 terms; $1 ; 6 ; 11$ or $7 ; 10 ; 13$.