? GP and Function Relation

Given: \( f(x) = x^3 + 3x - 9 \)

The sum of infinite GP = max value of \( f(x) \) on [−2, 3]

The difference between first two terms = \( f'(0) \)

Step 1: \( f(x) \) is increasing ⇒ Max at \( x = 3 \)

\( f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27 \)

Step 2: \( f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3 \)

⇒ \( a(1 - r) = 3 \)

Step 3: Solve:
\( a = 27(1 - r) \)
\( \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3} \)

✅ Final Answer: \( r = \frac{2}{3} \)