1B. Does there exist a real number $y$ such that the numbers $\sqrt{y^{2}+2 y+1}, \frac{y^{2}+3 y-1}{3}$, $y-1$ in the given order, are three consecutive terms of an arithmetic progression? Explain your answer.
Solution. If $p, q, r$ are three consecutive terms of an arithmetic progression, then the sum of the first and third term is twice the value of the second term, i.e., $p+r=2q$. In our case, this condition takes the form

$$
\sqrt{y^{2}+2 y+1}+y-1=2 \frac{y^{2}+3 y-1}{3} \text { or } 3\left(\sqrt{(y+1)^{2}}+y-1\right)=2\left(y^{2}+3 y-1\right) \text {. }
$$

Since $\sqrt{(y+1)^{2}}=|y+1|$, we get $3(|y+1|+y-1)=2\left(y^{2}+3 y-1\right)$. If we simplify the last equation, we get

$$
2 y^{2}+3 y-3|y+1|+1=0
$$

Equation (1) is equivalent to the system

$$
\left\{\begin{array} { l } 
{ 2 y ^ { 2 } + 3 y - 3 ( y + 1 ) + 1 = 0 } \\
{ y + 1 \geq 0 }
\end{array} \text { or } \left\{\begin{array}{l}
2 y^{2}+3 y+3(y+1)+1=0 \\
y+1<0
\end{array}\right.\right.
$$

from which, after simplification, they can be written in the form

$$
\left\{\begin{array} { l } 
{ y ^ { 2 } - 1 = 0 } \\
{ y \geq - 1 }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
y^{2}+3 y-2=0 \\
y<-1
\end{array}\right.\right.
$$

The solution to the first system is $y=1$ or $y=-1$, and the solution to the second system is $y=-2$.

Therefore, the numbers $y$ for which $\sqrt{y^{2}+2 y+1}, \frac{y^{2}+3 y-1}{3}, y-1$ are consecutive terms of an arithmetic progression are $y \in\{-2,-1,1\}$.