This problem involves finding the common ratio of a decreasing geometric progression (GP) by using information derived from a given function and its derivative.

Finding the Maximum Value of the Function

First, we need to find the greatest value of the function $f(x) = x^3 + 3x - 9$ in the interval $[-2, 3]$.

1. Calculate the derivative of the function: The derivative, $f'(x)$, helps us find critical points and understand the function's behavior. $f'(x) = \frac{d}{dx}(x^3 + 3x - 9) = 3x^2 + 3$.
2. Analyze the derivative: To find the maximum or minimum values within an interval, we look for critical points where $f'(x) = 0$. Set $f'(x) = 0$: $3x^2 + 3 = 0$ $3x^2 = -3$ $x^2 = -1$ This equation has no real solutions for $x$. This means there are no critical points within the real number line.
3. Evaluate the function at the interval endpoints: Since $f'(x) = 3x^2 + 3$ is always positive for all real $x$ (because $x^2 \ge 0$, so $3x^2 + 3 \ge 3$), the function $f(x)$ is strictly increasing throughout its domain, including the interval $[-2, 3]$. Therefore, the greatest value will occur at the right endpoint of the interval. Evaluate $f(x)$ at $x = -2$: $f(-2) = (-2)^3 + 3(-2) - 9 = -8 - 6 - 9 = -23$. Evaluate $f(x)$ at $x = 3$: $f(3) = (3)^3 + 3(3) - 9 = 27 + 9 - 9 = 27$.
4. Identify the greatest value: Comparing the values, the greatest value of $f(x)$ in the interval $[-2, 3]$ is $27$.

Calculating the Value of f'(0)

Next, we need to calculate the value of the derivative at $x=0$, which is given as the difference between the first two terms of the GP.

1. Substitute $x=0$ into the derivative: $f'(x) = 3x^2 + 3$ $f'(0) = 3(0)^2 + 3 = 0 + 3 = 3$.

So, the difference between the first two terms of the GP is $3$.

Relating GP Properties to Function Values

Let the first term of the decreasing GP be $a$ and the common ratio be $r$. For a decreasing GP, we must have $a > 0$ and $0 < r < 1$ (or $a < 0$ and $-1 < r < 0$). Since the sum is positive ($27$), $a$ must be positive, confirming $a > 0$ and $0 < r < 1$.

• Sum of infinite terms: The sum ($S$) of an infinite decreasing GP is given by the formula $S = \frac{a}{1-r}$. We know the sum is equal to the greatest value of the function, which is $27$. So, $\frac{a}{1-r} = 27$. Let's call this Equation (1).
• Difference between the first two terms: The first term is $a$ and the second term is $ar$. The difference is $a - ar = a(1-r)$. We know this difference is equal to $f'(0)$, which is $3$. So, $a(1-r) = 3$. Let's call this Equation (2).

Solving for the Common Ratio (r)

Now we have a system of two equations with two variables ($a$ and $r$):

1. $\frac{a}{1-r} = 27 \implies a = 27(1-r)$
2. $a(1-r) = 3$

Substitute the expression for $a$ from Equation (1) into Equation (2):

$$[27(1-r)](1-r) = 3$$ $$27(1-r)^2 = 3$$

Divide both sides by $27$:

$$(1-r)^2 = \frac{3}{27}$$ $$(1-r)^2 = \frac{1}{9}$$

Take the square root of both sides:

$$1-r = \pm \sqrt{\frac{1}{9}}$$ $$1-r = \pm \frac{1}{3}$$

We have two possible cases:

• Case 1: $1-r = \frac{1}{3}$ $r = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$
• Case 2: $1-r = -\frac{1}{3}$ $r = 1 - (-\frac{1}{3}) = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3}$

The problem states that the GP is a *decreasing* GP. This condition requires that the absolute value of the common ratio $|r|$ must be less than 1.

• In Case 1, $r = \frac{2}{3}$. Here, $|r| = |\frac{2}{3}| = \frac{2}{3} < 1$. This is a valid common ratio for a decreasing GP.
• In Case 2, $r = \frac{4}{3}$. Here, $|r| = |\frac{4}{3}| = \frac{4}{3} > 1$. This is not a valid common ratio for a decreasing GP.

Therefore, the common ratio of the decreasing GP is $\frac{2}{3}$.

Final Answer Verification

Let's check if $a > 0$ when $r = \frac{2}{3}$. Using Equation (2): $a(1 - \frac{2}{3}) = 3 \implies a(\frac{1}{3}) = 3 \implies a = 9$. Since $a = 9$ is positive and $r = \frac{2}{3}$ is between $0$ and $1$, the conditions for a decreasing GP are met.

The sum is $S = \frac{a}{1-r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27$. This matches the maximum value of the function.

The difference between the first two terms is $a(1-r) = 9(1 - 2/3) = 9(1/3) = 3$. This matches $f'(0)$.