In a geometric sequence, the $n$-th term is given by the formula:

$a_n = a \cdot r^{n-1}$

Where:

• $a$ is the first term
• $r$ is the common ratio

Given:

• The third term, $a_3 = 9/2$
• The sixth term, $a_6 = -2/3$

We can write these equations as:

1. $a \cdot r^2 = \frac{9}{2}$
2. $a \cdot r^5 = -\frac{2}{3}$

Step 1: Divide the equations to eliminate $a$

Divide the second equation by the first:

$\frac{a \cdot r^5}{a \cdot r^2} = \frac{-\frac{2}{3}}{\frac{9}{2}}$

$r^{5-2} = \frac{-\frac{2}{3}}{\frac{9}{2}}$

$r^3 = -\frac{2}{3} \cdot \frac{2}{9}$

$r^3 = -\frac{4}{27}$

Step 2: Solve for $r$

Take the cube root of both sides:

$r = \sqrt[3]{-\frac{4}{27}}$

$r = -\frac{\sqrt[3]{4}}{3}$

Step 3: Substitute $r$ into the first equation to find $a$

Using $a \cdot r^2 = \frac{9}{2}$, substitute $r = -\frac{\sqrt[3]{4}}{3}$:

$a \cdot \left(-\frac{\sqrt[3]{4}}{3}\right)^2 = \frac{9}{2}$

$a \cdot \frac{\sqrt[3]{16}}{9} = \frac{9}{2}$

Solve for $a$:

$a = \frac{\frac{9}{2}}{\frac{\sqrt[3]{16}}{9}}$

$a = \frac{9}{2} \cdot \frac{9}{\sqrt[3]{16}}$

$a = \frac{81}{2\sqrt[3]{16}}$

Thus, the first term