4.013. 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 10 terms of the arithmetic progression.
## Solution.

From the condition, we have: $a_{1}-1, a_{1}+d-1, a_{1}+2 d+1 \cdots$ - three consecutive terms of a geometric progression. Using formula (4.5), we find $S_{3}=\frac{2 a_{1}+2 d}{2} \cdot 3=15$ or $a_{1}+d=5$. Using formula (4.7), we get $\left(a_{1}+d-1\right)^{2}=\left(a_{1}-1\right)\left(a_{1}+2 d+1\right)$. Substituting the value $a_{1}=5-d$ into this equation, we obtain $16=(4-d)(6+d), d^{2}+2 d-8=0$. From this, $d_{1}=-4$, $d_{2}=2$. Then $a_{1}^{\prime}=9, a_{2}^{\prime}=5, a_{3}^{\prime}=1 ; a_{1}^{\prime \prime}=3, a_{2}^{\prime \prime}=5, a_{3}^{\prime \prime}=7$. Considering that by the condition $a_{1}<a_{2}<a_{3}$, we get $a_{1}=3, d=2$. Then

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

Answer: 120.