$x$, $y$, and $\frac3{2x}$ are non-zero terms in an arithmetic progression. If the third term is increased by $1$, the three terms now form a geometric progression. Find the original three terms.
Here are my steps:
1) The numbers were set into a proportion to find the common ratio. $\dfrac xy = \dfrac{\tfrac3{2x}}y$
2) The equation was set equal to $0$ such that $0=\tfrac32x^2-y^2$
I do not know how to move on from here or if the steps lead to the correct answer. How would you solve this problem?