polynomial

Let the three roots be $a$ , $b$ , and $c$ such that $a < b < c$ . Since they are in a arithmetic progression, $a+c= 2b$ . By Vieta's formula , we have:

$\begin{aligned} a + b + c & = \frac {144}{64} \\ 3b & = \frac 94 \\ \implies b & = \frac 34 \end{aligned}$

By Vieta's formula again,

$\begin{aligned} a b c & = \frac {15}{64} \\ (b-\blue d)b(b+\blue d) & = \frac {15}{64} & \small \blue{\text{where }d \text{ is the common difference.}} \\ \frac 34 \left(\frac 34-d\right)\left(\frac 34+d\right) & = \frac {15}{64} & \small \blue{\text{Note that }b = \frac 34} \\ \frac 9{16} - d^2 & = \frac 5{16} \\ d^2 & = \frac 14 & \small \blue{\text{Since }d > 0} \\ \implies d & = \frac 12 \end{aligned}$

Then the largest root $c = b+d = \dfrac 34 + \dfrac 12 = \dfrac 54$ and $p+q = 5+4 = \boxed 9$ .

Since the roots are in APRIL, we can express them as $a-d,a,a+d$

The sum of the roots is $\frac{-b}{2a}$ , which is $\frac{144}{64}$ . Adding up the roots will leave only $a$ 's, and in turn, $a=\frac{3}{4}$

The product of the roots of any odd degree polynomial is $\frac{-d}{a}$ , which is $\frac{15}{64}$ . Multiplying the roots with the found value of $a$ will go as follows:

$\frac{3}{4}(\frac{3}{4}-d)(\frac{3}{4}+d)=\frac{15}{64}$

Solving the equation will give us $d=\frac{1}{2}$

The biggest roots is $a+d$ which is $\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$

And so the answer is $\boxed{9}$