Correct Answer - Option 3 :
\(\frac 2 3\)
Concept:
GP is represented as \(a,ar,a{r^2},a{r^3},............,a{r^n}\), where a is the first term, r is the common ratio.
The sum of infinite terms of a decreasing GP\( = \frac{a}{{1 - r}}\)
Calculation:
f(x) = x3 + 3x - 9
\(f'(x) = 3{x^2} + 3 = 3({x^2} + 1) \ge 3\) for all x ∈ R
f(x) is strictly increasing function for all x ∈ R
so the greatest value of f(x) in the interval is [-2, 3] is at x = 3 So, let's find f(3)
⇒ f(3) = 27
According to question we know that the sum of infinite terms of a decreasing GP is equal to the greatest value of the function f(x)
\(⇒ \frac{a}{{1 - r}} = 27\)
⇒ a = 27 × (1 - r) -----------(1)
According to question, the difference between the first two terms is f'(0)
⇒ a - ar = f'(0) = 3
⇒ a(1 - r) = 3 ------------(2)
So, using equation (1) and (2) we get,
⇒ 27 × (1 - r) × (1 - r) = 3
⇒ (1 - r)2 = 1/9
⇒ (1 - r) = ± 1/3
\(\Rightarrow r = \frac{2}{3}\ or\ \frac{4}{3}\)
GP is decreasing series so common ratio \(r = \frac{2}{3}\)