Question 1000064
In general, the answer is NONE.
Adding some requirements to the coefficients a, b, c, and d,
we could be sure to have a gp,
or even a gp that is also an ap, and a hp.


If there are no specifications about a,b,c,and d,
{{{ax^3+bx^2+cx+d}}} could be divisible by {{{ax^2+c}}} if {{{system(a=b,c=d=0)}}} ,
and then a,b,c,d would be neither an ap, nor a gp, nor a hp.


If it were specified that none of the coefficients a, b,c,and d is zero,
when we divide {{{ax^3+bx^2+cx+d}}} by {{{ax^2+c}}} we find
a quotient of {{{x+b/a}}} and
a remainder of {{{d-bc/a}}} .
In that case,
if {{{ax^3+bx^2+cx+d}}} is divisible by {{{ax^2+c}}} , then
{{{d-bc/a=0}}}--->{{{d=bc/a}}}--->{{{d/b=c/a}}} .
Conversely, if {{{d/b=c/a}}}<--->{{{d=bc/a}}}<--->{{{ad=bc}}}<--->{{{d/c=b/a}}} ,
then the remainder is zero, and {{{ax^3+bx^2+cx+d}}} is divisible by {{{ax^2+c}}} .
For {{{d/b=c/a}}}<--->{{{d=bc/a}}} to happen,
it is not necessary that a,b,c,d form an ap, an gp, or an hp.
Sometimes a,b,c,d will be an ap, a gp, and a hp,
sometimes it will only be a gp,
but other times it will be none of those.
For example, 1,2,3,6 is neither an ap, nor a gp, nor a hp.


With {{{d/b=c/a}}}<--->{{{d=bc/a}}} , and {{{b=sqrt(ac)}}}<--->{{{b/a=sqrt(c/a)}}} ,
a,b,c,d is a gp with common ratio {{{sqrt(c/a)}}} .
1,2,4,8 is an example.
A special case of that is when {{{a=b=c=d}}} , which makes a,b,c,d
a gp, an ap, and a hp.