6.17 The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 189. Find the first term and the common ratio of this progression.
6.17 According to the condition, we have

$$
\left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}\right)=21 \\
b_{1}^{2}\left(1+q^{2}+q^{4}\right)=189
\end{array}\right.
$$

Squaring both sides of the first equation:

$$
b_{1}^{2}\left(1+q^{2}+q^{4}\right)+2 b_{1}^{2} q\left(1+q+q^{2}\right)=441
$$

Subtracting the second equation from this equation, we get

$$
2 b_{1}^{2} q\left(1+q+q^{2}\right)=252
$$

from which $b_{1} q=6$. Now from the system

$$
\left\{\begin{array}{l}
b_{1}\left(1+q^{2}\right)=15 \\
b_{1} q=6
\end{array}\right.
$$

we find two solutions: 1) $q=2, b_{1}=3$; 2) $q=\frac{1}{2}, b_{1}=12$. Checking shows that both solutions are valid.

Answer: $b_{1}=3, q=2$ or $b_{1}=12, q=\frac{1}{2}$.