Question

Sequences and Their Limits. An infinitely Decreasing Geometric Progression. Limits of Functions.
The sum of the terms of an infinitely decreasing geometric progression is equal to the greatest value of the function $f\left(x\right)=x^3+3x-9$ on the interval $[-2,3]$; the difference between the first and the second terms of the progression is $f^{\prime }\left(0\right)$. Find the common ratio of the progression.

Answer 1

Here maximum value of $f\left(x\right)=x^3+3x-9$ in interval $[-2/3]$

will be at $x=3f\left(3\right)=27+9-9$

$=27$

$\Rightarrow$ here maximum value $=$ Sum of infinite term of G.P.

$\Rightarrow 27=\frac{a}{1-r}.....\left(i\right)$

$a=firstterm$

$v=commonratio$

and difference b/w two terms $a-ar=f^{\prime }\left(0\right)$

$\Rightarrow a-ar=(3x^2+3)x=0$

$\Rightarrow a-ar=3.....\left(ii\right)$

divide $e{q}^n$ (i) and (ii)

So, here $\frac{27}{3}=\frac{a}{\frac{1-r}{a(1-r)}}\Rightarrow 9=\frac{1}{(1-r{)}^2}$

$\Rightarrow \frac{1}{1-r}=3$

$\Rightarrow 3-3r=1$

$\Rightarrow 2=3r$

$\Rightarrow 1=2/3\Rightarrow Commonration=2/3$.