6.23 The sum of the first three terms of an increasing arithmetic progression is 21. If 1 is subtracted from the first two terms of this progression, and 2 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first eight terms of the geometric progression.
6.23 Let $a_{1}$ be the first term of an arithmetic progression, and $d$ its difference. Then, according to formula (6.2), we have $S_{3}=\frac{2 a_{1}+2 d}{2} \cdot 3=21$, or $a_{1}+d=7$. By the condition, $a_{1}-1$, $a_{1}+d-1, a_{1}+2 d+2$ are three consecutive terms of a geometric progression. Using formula (6.8), we get

$$
\left(a_{1}+d-1\right)^{2}=\left(a_{1}+2 d+2\right)\left(a_{1}-1\right)
$$

from which, after substituting $a_{1}=7-d$ and expanding the brackets, we arrive at the quadratic equation $d^{2}+3 d-18=0$, and thus $d_{1}=3, d_{2}=-6$. Only $d_{1}=3$ satisfies the condition; then $a_{1}=4$. Next, we find $b_{1}=a_{1}-1=3$, $b_{2}=a_{1}+d-1=6$, and therefore, $q=2$. Finally, using formula (6.7a), we get

$$
S_{8}=\frac{a_{1}\left(q^{8}-1\right)}{q-1}=\frac{3\left(2^{8}-1\right)}{2-1}=765
$$

Answer: 765.