The sum of the terms of an infinitely decreassing Geometric Progression (GP) is equal to the greatest value of the function `f (x) = x^(3) + 3x -9` where `x in [-4, 3] and ` the difference between the first and second term is `f'(0).` The common ratio `r = p/q` where p and q are relatively prime positive integers. Find `(p+q).`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and derive the necessary values systematically.

### Step 1: Define the Geometric Progression (GP)
Let the first term of the GP be \( a \) and the common ratio be \( r \). The terms of the GP are:
- First term: \( a \)
- Second term: \( ar \)

### Step 2: Sum of the Infinite GP
The sum \( S \) of an infinitely decreasing GP is given by the formula:
\[ S = \frac{a}{1 - r} \]
According to the problem, this sum is equal to the greatest value of the function \( f(x) \).

### Step 3: Find the Maximum Value of \( f(x) \)
The function given is:
\[ f(x) = x^3 + 3x - 9 \]
To find the maximum value of \( f(x) \) on the interval \( [-4, 3] \), we first need to find the derivative \( f'(x) \):
\[ f'(x) = 3x^2 + 3 \]
Since \( f'(x) \) is always positive (as \( 3x^2 + 3 > 0 \) for all \( x \)), the function \( f(x) \) is increasing on the entire interval. Therefore, the maximum value occurs at the right endpoint \( x = 3 \):
\[ f(3) = 3^3 + 3 \cdot 3 - 9 = 27 + 9 - 9 = 27 \]

### Step 4: Set Up the Equation for the Sum of the GP
From the previous steps, we have:
\[ \frac{a}{1 - r} = 27 \quad \text{(Equation 1)} \]

### Step 5: Find the Difference Between the First and Second Terms
The difference between the first and second terms of the GP is:
\[ a - ar = a(1 - r) \]
According to the problem, this difference is equal to \( f'(0) \). We calculate \( f'(0) \):
\[ f'(0) = 3(0)^2 + 3 = 3 \]
Thus, we have:
\[ a(1 - r) = 3 \quad \text{(Equation 2)} \]

### Step 6: Solve the System of Equations
From Equation 1, we can express \( a \):
\[ a = 27(1 - r) \]
Substituting this into Equation 2:
\[ 27(1 - r)(1 - r) = 3 \]
\[ 27(1 - r)^2 = 3 \]
Dividing both sides by 27:
\[ (1 - r)^2 = \frac{3}{27} = \frac{1}{9} \]
Taking the square root:
\[ 1 - r = \frac{1}{3} \quad \text{or} \quad 1 - r = -\frac{1}{3} \]
Solving these gives:
1. \( 1 - r = \frac{1}{3} \) → \( r = \frac{2}{3} \)
2. \( 1 - r = -\frac{1}{3} \) → \( r = \frac{4}{3} \)

Since we need an infinitely decreasing GP, we discard \( r = \frac{4}{3} \) (as it is greater than 1).

### Step 7: Identify \( p \) and \( q \)
Thus, we have:
\[ r = \frac{2}{3} \]
Here, \( p = 2 \) and \( q = 3 \).

### Step 8: Calculate \( p + q \)
Finally, we find:
\[ p + q = 2 + 3 = 5 \]

### Final Answer
The value of \( p + q \) is \( \boxed{5} \).

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