Question

The sum of an infinitely decreasing geometric progression whose first term is a and common ratio $r$, is equal to least value of the quadratic trinomial $P ( x )=3 x ^2- x +\frac{25}{12}$, in [0,2]. Also the first term of the geometric progression is equal to the square of its common ratio.
The value of $\left(r^2+2 r\right)$ is

• A. 1
• B. 2
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

Given, $ P(x)=3 x^2-x+\frac{25}{12}=3\left[x^2-\frac{1}{3} x+\frac{25}{36}\right]=3\left(\left(x-\frac{1}{6}\right)^2+\frac{2}{3}\right)$
$\left.\Rightarrow P ( x )\right|_{\min .}=2$, occurs at $x =\frac{1}{6}$
Given, $\frac{ a }{1- r }=2$....(1)
and $a=r^2$
$\therefore $ Solving (1) and (2), we get
$\frac{r^2}{1-r}=2 \Rightarrow r^2=2-2 r \Rightarrow r^2+2 r=2$