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The sum of infinite terms of decreasing GP is equal to the greatest value of the function

$f(x) = x^3 + 3x – 9$ in the interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is

1
2
3
4

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: $r = \frac{2}{3}$

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