6.15 The sum of the first three terms of an increasing arithmetic progression is 15. If 1 is subtracted from the first two terms of this progression, and 1 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first ten terms of the arithmetic progression.
6.15 Let $a_{1}$ be the first term of an arithmetic progression, and $d$ its difference. According to the condition, $\frac{\left(2 a_{1}+2 d\right) 3}{2}=15$, from which $a_{1}+d=5$.

Further, since the numbers $a_{1}-1, a_{1}+d-1, a_{1}+2 d+1$ form a geometric progression, by formula (6.8) we have

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

Solving the system of equations (1), (2), we get two solutions: 1) $\left.a_{1}=3, d=2 ; 2\right) a_{1}=9, d=-4$. However, the second solution does not fit $(d=-4<0)$. Therefore,

$$
S_{10}=\frac{2 \cdot 3+2 \cdot 9}{2} \cdot 10=120
$$

Answer: 120.