4.027. The sum of the third and ninth terms of an arithmetic progression is 6, their product is $135 / 16$. Find the sum of the first 15 terms of this progression.
## Solution.

From the condition, we have $\left\{\begin{array}{l}a_{3}+a_{9}=6, \\ a_{3} \cdot a_{9}=\frac{135}{16}\end{array}\right.$.

Using formula (4.1), we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ a _ { 1 } + 2 d + a _ { 1 } + 8 d = 6 , } \\
{ ( a _ { 1 } + 2 d ) ( a _ { 1 } + 8 d ) = \frac { 1 3 5 } { 1 6 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{1}+5 d=3, \\
\left(a_{1}+2 d\right)\left(a_{1}+8 d\right)=\frac{135}{16}
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow a_{1}=3-5 d \Rightarrow(3-5 d+2 d)(3-5 d+8 d)=\frac{135}{16}, \\
& d^{2}=\frac{1}{16}, d_{1}=-\frac{1}{4}, d_{2}=\frac{1}{4} . \\
& \text { Then } a_{1}^{\prime}=\frac{17}{4} \text { and } a_{1}^{\prime \prime}=\frac{7}{4} .
\end{aligned}
$$

Using formula (4.5), we get

$$
S_{15}^{\prime}=\frac{\frac{17}{2}-\frac{1}{4} \cdot 14}{2} \cdot 15=37.5 \text { or } S_{15}^{\prime \prime}=\frac{\frac{7}{2}+\frac{1}{4} \cdot 14}{2} \cdot 15=52.5
$$

Answer: 37.5 or 52.5.