4.009. Find the third term of an infinite geometric progression with a common ratio $|q|<1$, the sum of which is equal to $8 / 5$, and the second term is $-1 / 2$.
Solution.

Using the formulas $S=\frac{b_{1}}{1-q}$ and $b_{n}=b_{1} q^{n-1}$, we get $\left\{\begin{array}{l}\frac{b_{1}}{1-q}=\frac{8}{5}, \\ b_{1} q=-\frac{1}{2}\end{array} \Rightarrow\right.$
$\Rightarrow b_{1}=\frac{8-5 q}{5}, 16 q^{2}-16 q-5=0$, from which we find $q_{1}=-\frac{1}{4}, q_{2}=\frac{5}{4}>1$ (does not fit). Then $b_{3}=b_{1} q^{2}=\left(-\frac{1}{2}\right) \cdot\left(-\frac{1}{4}\right)=\frac{1}{8}$.

Answer: $\frac{1}{8}$.