4.002. The sum of the first and fifth terms of an arithmetic progression is 5/3, and the product of the third and fourth terms is 65/72. Find the sum of the first 17 terms of this progression.
Solution.

We have $\left\{\begin{array}{l}a_{1}+a_{5}=\frac{5}{3}, \\ a_{3} \cdot a_{4}=\frac{65}{72} .\end{array}\right.$

Using formula (4.1), we find

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ a _ { 1 } + a _ { 1 } + 4 d = \frac { 5 } { 3 } , } \\
{ ( a _ { 1 } + 2 d ) ( a _ { 1 } + 3 d ) = \frac { 6 5 } { 7 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 a_{1}+4 d=\frac{5}{3}, \\
\left(a_{1}+2 d\right)\left(a_{1}+3 d\right)=\frac{65}{72}
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } + 2 d = \frac { 5 } { 6 } , } \\
{ ( a _ { 1 } + 2 d ) ( ( a _ { 1 } + 2 d ) + d ) = \frac { 6 5 } { 7 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}+2 d=\frac{5}{6}, \\
\frac{5}{6}\left(\frac{5}{6}+d\right)=\frac{65}{72}
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } = \frac { 5 } { 6 } - 2 d , } \\
{ d = \frac { 1 } { 4 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=\frac{5}{6}-2 \cdot \frac{1}{4}=\frac{1}{3}, \\
d=\frac{1}{4}
\end{array}\right.\right.
\end{aligned}
$$

Using formula (4.5), we get $S_{17}=\frac{2 \cdot \frac{1}{3}+\frac{1}{4} \cdot 16}{2} \cdot 17=\frac{119}{3}$.

Answer: $\frac{119}{3}$.