Find the three geometric means between 128 and 40 rac12.

2. Find the three geometric means between 128 and $40\frac{1}{2}$.

Step 1: Identify the terms and set up the equation.
Let the first term be $a = 128$. The last term is $40\frac{1}{2} = \frac{81}{2}$.
If there are three geometric means ($G_1, G_2, G_3$) between 128 and $\frac{81}{2}$, the sequence is $128, G_1, G_2, G_3, \frac{81}{2}$.
This means there are $n=5$ terms in total. The 5th term is $a_5 = \frac{81}{2}$.
The formula for the $n$-th term of a geometric progression is $a_n = ar^{n-1}$.
So, for the 5th term:

$$a_5 = ar^{5-1} = ar^4$$ $$\frac{81}{2} = 128r^4$$

Step 2: Solve for the common ratio $r$.
Divide both sides by 128:

$$r^4 = \frac{81}{2 \times 128}$$ $$r^4 = \frac{81}{256}$$

Take the fourth root of both sides:

$$r = \sqrt[4]{\frac{81}{256}}$$ $$r = \frac{\sqrt[4]{81}}{\sqrt[4]{256}}$$ $$r = \frac{3}{4}$$

(We consider the positive root for geometric means unless specified otherwise, as typically geometric means are positive when the terms are positive).

Step 3: Calculate the three geometric means.
The geometric means are $G_1 = ar$, $G_2 = ar^2$, and $G_3 = ar^3$.

$$G_1 = 128 \times \frac{3}{4} = 32 \times 3 = 96$$ $$G_2 = 128 \times \left(\frac{3}{4}\right)^2 = 128 \times \frac{9}{16} = 8 \times 9 = 72$$ $$G_3 = 128 \times \left(\frac{3}{4}\right)^3 = 128 \times \frac{27}{64} = 2 \times 27 = 54$$

The three geometric means are 96, 72, 54.

$$\boxed{\text{96, 72, 54}}$$

3. The sum of the first $n$ terms of a sequence is given by $S_n = \frac{3n^2}{2} - \frac{3n}{2}$.