If the sum of the terms of an infinitely decreasing GP is equal to the greatest value of the fuction `f(x)=x^(3)+3x-9` on the iterval `[-5,3]` and the difference between the first and second terms is `f'(0)` , then show that the common ratio of the progression is `(2)/(3)`.

Explanation

Given, `f(x)=x^(3)+3x-9`
`:. F'(x)=3x^(2)+3`
Hence, `f'(x)gt0` in `[-5,3]` then `f(x)` is an increasing function on `[-5,3]` and `f(x)` will have greatest value at `x=3`.
Thus, greatest value of `f(x)` is
`f(x)=3^(3)+3*3-9=27`
Let `a,ar,ar6(2),"......"` be a GP with common ratio `|r|lt1 {:. " given infinitely GP "]`
and also given `S_(oo)=27`
so, `(a)/(1-r)=27 " " "......(i)"`
and `a-ar=f'(0)`
`implies a(1-r)=f'(0)=3" " " " [:. f'(0)=3]`
`:. a(1-r)=3 " " ".....(ii)"`
From Eqs.(i) and (ii), we get
`(1-r)^(2)=(1)/(9) implies 1-r=pm(1)/(3)`
`:.r=1pm(1)/(3)`
So, `r=(4)/(3),(2)/(3) implies r ne (4)/(3) " " [:.|r|lt1]`
Hence, `r=(2)/(3)`