assume the contrary that there exist geometric progressions $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ such that: $a_1=b_1+c_1,a_2=b_2+c_2,a_3=b_3+c_3,a_4=b_4+c_4.$ we can easily check that $0\notin\{b_1,c_1\}$ and $0\notin\{x,y\}$ let $x,y$ be the common ratio of $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ respectively thus $b_1=\frac{b_2}{x}=\frac{b_3}{x^2}=\frac{b_4}{x^3}$ and $c_1=\frac{c_2}{y}=\frac{c_3}{y^2}=\frac{c_4}{y^3}$ since $a_2-a_1=a_3-a_2=a_4-a_3$ $\Rightarrow (b_2+c_2)-(b_1+c_1)=(b_3+c_3)-(b_2+c_2)=(b_4+c_4)-(b_3+c_3)$ $\Rightarrow b_1(x-1)+c_1(y-1)=b_1x(x-1)+c_1y(y-1)=b_1x^2(x-1)+c_1y^2(y-1)$ therefore $b_1(x-1)^2+c_1(y-1)^2=0~~~(*)$ and $b_1x(x-1)^2+c_1y(y-1)^2=0~~~~~(**)$ from $(*)\Rightarrow b_1(x-1)^2=-c_1(y-1)^2$ thus $-c_1x(y-1)^2=c_1y(y-1)^2~~;~from~(**)$ $\because$ $0\notin\{b_1,c_1\}$ and $0\notin\{x,y\}~\therefore~y=1 $ in a similar way we'll get that $x=1$ therefore $a_1=a_2=a_3=a_4$ which contradicts with $d\ne0$ $\therefore$ we can conclude that there are no geometric progressions $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ such that: $a_1=b_1+c_1,a_2=b_2+c_2,a_3=b_3+c_3,a_4=b_4+c_4.$ as desired $\square$