Let $f(x)=\int_0^x(4t^4-at^3)dt$ and $g(x)$ be a quadratic polynomial satisfying $g(0)+6=g'(0)-c=g''(0)+2b=0$, where $a,b,c$ are positive real numbers 
Let $f(x)=\int_0^x(4t^4-at^3)dt$ and $g(x)$ be a quadratic polynomial satisfying $g(0)+6=g'(0)-c=g''(0)+2b=0$, where $a,b,c$ are positive real numbers. If $y=g(x)$ and $y=f'(x)$ intersects in $4$ distinct points with abscissae $x_i=1,2,3,4$ such that $\sum_{i=1}^4\frac i{x_i}=8$; then are the abscissae of the point of intersection in Arithmetic Progression or Geometric Progression or Harmonic Progression?

Let $g(x)=px^2+qx+r$
$g(0)+6=0\implies r=-6$
$g'(0)-c=0\implies q=c$
$g''(0)=2b\implies p=-b$
$\implies g(x)=-bx^2+cx-6$
$f'(x)=4x^4-ax^3$
$f'(x)=g(x)\implies4x^4-ax^3+bx^2-cx+6=0$ has four roots $x_1,x_2,x_3,x_4$
Given, $\frac1{x_1}+\frac2{x_2}+\frac3{x_3}+\frac4{x_4}=8$
I tried to use Vieta's formula and got equations, but not able to reach anywhere.
In the hint, it's written Harmonic Mean=Geometric Mean. I don't understand why and even how to use that to get to the answer.
 A: I will pick up from where you left off.
$$\frac1{x_1}+\frac2{x_2}+\frac3{x_3}+\frac4{x_4}=8\qquad\ldots(1)$$
Using $AM\geq GM\geq HM$ on $x_1,\frac{x_2}{2},\frac{x_3}{3},\frac{x_4}{4},$ we have
$$\Rightarrow \frac{\sum_{i=1}^4\frac{x_i}{i}}{4}\geq \left(\frac{x_1x_2x_3x_4}{24}\right)^{\frac14}\geq\frac{4}{\sum_{i=1}^4\frac{i}{x_i}}=\frac12\qquad\ldots(2)$$
From the equation $4x^4-ax^3+bx^2-cx+6=0$, we have $$\Rightarrow x_1x_2x_3x_4=\frac64=\frac32$$
Using this value in $(2)$, we get
$$\Rightarrow \left(\frac{x_1x_2x_3x_4}{24}\right)^{\frac14}=\left(\frac{\frac32}{24}\right)^\frac14=\frac12=\frac{4}{\sum_{i=1}^4\frac{i}{x_i}}$$
Hence, we have $GM=HM$ which will occur iff $x_1=\frac{x_2}{2}=\frac{x_3}{3}=\frac{x_4}{4}$
Now you can easily obtain the values of $x_1,x_2,x_3,x_4$ and check if they are in $AP$,$GP$ or $HP$.
A: Apply GM-HM inequality,
$$\frac 12=\left( \frac{6}{4\cdot 24}\right)^{1/4}=\left( \frac{x_1x_2x_3x_4}{24}\right)^{1/4}\ge\frac{4}{\frac1{x_1}+\frac2{x_2}+\frac3{x_3}+\frac4{x_4}}=\frac 12$$
So $\displaystyle \frac{x_1}{1}=\frac{x_2}{2}=\frac{x_3}{3}=\frac{x_4}{4}=\frac 12$
S0 $x_1,x_2,x_3,x_4$ are in AP.
