Let a, b, c be distinct positive numbers in Arithmetic progression such that their Geometric mean is $b^{2/3}andc>a+2$. If their common difference is d, where $d\in N$ is minimum, then the value of ${(2b-1)}^2=\frac{p}{q},$   where $p,q$ are natural numbers with HCF 2, then the value of $p-16q$ is __

$$\begin{gathered} Letb-d,b+d \\ \Rightarrow {[(b-d).b(b+d)]}^{\frac{1}{3}}=b^{\frac{2}{3}}\Rightarrow (b^2-d^2)b=b^2 \\ \Rightarrow b^2-b-d^2=0 \Rightarrow d^2=b^2-b \\ {d}^2={(b-\frac{1}{2})}^2-\frac{1}{4} \\ {d}^2>-\frac{1}{4} \end{gathered}$$

Possible values of  $d^{2}are\{0,1,4,9,\mathrm{.....}\}$

$$\begin{gathered} At d^2=0\Rightarrow a=b=c \\ At d^2=1\Rightarrow b-1,b,b+1butc-a=2 \\ At d^2=4\Rightarrow b-2,b,b+2butc-a=4>2 \\ Now, {(2b-1)}^2=4b^2-4b+1 \\ =4(b^2-b)+1=4d^2+1=17 \end{gathered}$$