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

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

Answer 1

Answer: (B)

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