Question

The roots of the equation $x^2+2(a-3)x+9=0$ lie between $-6$ and $1.$ If $2,h_1,h_2,\dots ,h_{20},\left[a\right],$ where $[.]$ represents the greatest integer function, are in harmonic progression, then the value of $10h_9$ is

Answer 1

Roots of $x^2+2(a-3)x+9=0$ lie between $-6$ and $1.$
$\left(I\right)$ $D\ge 0$
$\Rightarrow 4(a-3{)}^2-4\cdot 9\ge 0$
$\Rightarrow a\in (-\infty ,0]\cup [6,\infty )$

$\left(II\right)$ $-6<\frac{-b}{2a}<1$
$\Rightarrow -6<3-a<1$
$\Rightarrow a\in (2,9)$

$\left(III\right)$ $f(-6)>0$
$\Rightarrow 36-12(a-3)+9>0$
$\Rightarrow a<\frac{27}{4}$

$\left(IV\right)$ $f\left(1\right)>0$
$\Rightarrow 1+2(a-3)+9>0$
$\Rightarrow a>-2$

Taking intersection, $a\in [6,\frac{27}{4})$
$\left[a\right]=6$

Now, $2,h_1,h_2,\dots ,h_{20},6$ are in H.P.
$\Rightarrow \frac{1}{2},\frac{1}{h_1},\frac{1}{h_2},\dots ,\frac{1}{h_{20}},\frac{1}{6}$ are in A.P.
$\frac{1}{6}=\frac{1}{2}+21d$ where $d$ is the common difference of A.P.
$\Rightarrow d=-\frac{1}{63}$
$\therefore \frac{1}{h_9}=\frac{1}{2}+9\left(\frac{-1}{63}\right)=\frac{5}{14}$
$\Rightarrow h_9=\frac{14}{5}$