1. The problem states that the ratio $r$ is given by $r = \frac{a_3}{a_2} = \frac{a_2}{a_1}$ and that $a_2^2 = a_1 a_3$. 2. This implies a geometric progression where the middle term squared equals the product of the first and third terms. 3. From the ratio equality, we have $\frac{a_3}{a_2} = \frac{a_2}{a_1} = r$. 4. Multiplying both sides of $\frac{a_3}{a_2} = r$ by $a_2$ gives $a_3 = r a_2$. 5. Similarly, from $\frac{a_2}{a_1} = r$, multiplying both sides by $a_1$ gives $a_2 = r a_1$. 6. Substitute $a_2 = r a_1$ and $a_3 = r a_2 = r^2 a_1$ into $a_2^2 = a_1 a_3$: $$ (r a_1)^2 = a_1 (r^2 a_1) $$ 7. Simplify the left side: $r^2 a_1^2 = r^2 a_1^2$. 8. Both sides are equal, confirming the relation holds true. Final answer: The given relations confirm that $a_1, a_2, a_3$ form a geometric progression with common ratio $r$ such that $a_2^2 = a_1 a_3$ and $r = \frac{a_2}{a_1} = \frac{a_3}{a_2}$.