$\text{ If }\mathrm{a},\mathrm{b},\mathrm{c} \mathrm{are}\text{ in A.P then }\frac{a}{bc},\frac{1}{c},\frac{2}{b}\text{ will be in }$

$\text{ Since }a{x}^3+b{x}^2+cx+d\text{ is divisible by }a{x}^2+c,\text{ therefore, when }a{x}^3+b{x}^2+cx+d$

$\text{ is divided by }a{x}^2+c\text{ the remainder should be zero, Now when }$

$a{x}^3+b{x}^2+cx+d\text{ is divided by }a{x}^2+c,\text{ then the remainder is }\frac{bc}{a}-d$

$\begin{array}{l}\therefore \frac{bc}{a}-d=0\\ \Rightarrow bc=ad\\ \Rightarrow \frac{b}{a}=\frac{d}{c}\\ \text{ Hence, from this }\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{ are not necessarily in G.P. }\end{array}$