Question

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function \( f(x) = x^3 + 3x - 9 \) in the interval \([-2, 3]\), and the difference between the first two terms is \( f'(0) \). Then the common ratio of the GP is:

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{3} \)

Hint:

To solve for the common ratio in geometric progressions, use the given conditions and equations, including the sum of infinite terms and the difference between the first two terms.

Answer 1

The Correct Option is C

Step 1: Analyze the given GP. 
Let the first term of the GP be \( a \) and the common ratio be \( r \). The sum of an infinite decreasing GP is given by: \[ S_{\infty} = \frac{a}{1 - r}. \]

Step 2: Maximize the function \( f(x) = x^3 + 3x - 9 \) in the interval \([-2, 3]\). 
To find the greatest value of \( f(x) \), we first calculate the derivative of \( f(x) \): \[ f'(x) = 3x^2 + 3. \] Setting \( f'(x) = 0 \), we get: \[ 3x^2 + 3 = 0 \Rightarrow x^2 = -1. \] This equation has no real solutions. Thus, we need to evaluate \( f(x) \) at the endpoints of the interval \([-2, 3]\). 
\[ f(-2) = (-2)^3 + 3(-2) - 9 = -8 - 6 - 9 = -23, \] \[ f(3) = (3)^3 + 3(3) - 9 = 27 + 9 - 9 = 27. \] The maximum value of \( f(x) \) in the interval \([-2, 3]\) is 27, which occurs at \( x = 3 \).

Step 3: Use the given condition for the sum of the GP. 
We are told that the sum of the infinite terms of the GP is equal to the greatest value of \( f(x) \), i.e., \[ S_{\infty} = 27. \] Thus, \[ \frac{a}{1 - r} = 27. \]

Step 4: Use the given condition on the difference between the first two terms. 
The difference between the first two terms of the GP is given as \( f'(0) \). First, calculate \( f'(0) \): \[ f'(0) = 3(0)^2 + 3 = 3. \] Therefore, the difference between the first two terms is 3, i.e., \[ a(1 - r) = 3. \]

Step 5: Solve the system of equations. 
From the equation \( \frac{a}{1 - r} = 27 \), we have: \[ a = 27(1 - r). \] Substitute this into \( a(1 - r) = 3 \): \[ 27(1 - r)(1 - r) = 3 \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9}. \] Taking the square root of both sides: \[ 1 - r = \frac{1}{3} \Rightarrow r = \frac{2}{3}. \]

Step 6: Conclusion. 
Thus, the common ratio of the GP is \( \frac{2}{3} \), and the correct answer is (c).