Question
The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function $$f\left( x \right) = {x^3} + 3x - 9$$ in the interval $$\left[ { - 2,3} \right]$$ and the difference between the first two terms is $$f'\left( 0 \right).$$ Then the common ratio of the G.P. is
A.
$$ - \frac{2}{3}$$
B.
$$\frac{4}{3}$$
C.
$$\frac{2}{3}$$
D.
$$ - \frac{4}{3}$$
Answer :
$$\frac{2}{3}$$
Solution :
Let the G.P. be $$a,ar,a{r^2},.....\left( {0 < r < 1} \right).$$ From the question, $$\frac{a}{{1 - r}} = {3^3} + 3.3 - 9$$
$$\eqalign{
& \left\{ {\because \,\,f'\left( x \right) = 3{x^2} + 3 > 0;\,{\text{so, }}f\left( x \right)\,{\text{is monotonically increasing;}}} \right. \cr
& \left. {\therefore \,\,f\left( 3 \right)\,{\text{is the greatest value in }}\left[ { - 2,3} \right].} \right\} \cr
& {\text{Also, }}f'\left( 0 \right) = 3.\,\,{\text{So, }}a - ar = 3. \cr
& {\text{Solving, }}a = 27\left( {1 - r} \right)\,{\text{and }}a\left( {1 - r} \right) = 3 \cr
& {\text{we get }}r = \frac{2}{3},\frac{4}{3}. \cr
& {\text{But }}r < 1. \cr} $$