Given that $a,b,c,d$ are in G.P.
Let $a,b,c,d$ represent first four terms of a G.P.
$\Rightarrow\:a=a,\:\:b=a.r,\:\:c=a.r^2,\:\:d=a.r^3$
To show three terms $x,y,z$ are in G.P. we have to show that $y^2=xy$
$\therefore$ To show that $(a^n+b^n),(b^n+c^n),(c^n+d^n)$ are in G.P we have
to show that $(b^n+c^n)^2=(a^n+b^n). (c^n+d^n)$
Step 2
$a^n+b^n=a^n+(a.r)^n=a^n(1+r^n)$
$b^n+c^n=(ar)^n+(ar^2)^n=a^nr^n(1+r^n)$
$c^n+d^n=(ar^2)^n+(ar^3)^n=a^n.r^{2n}(1+r^n)$
Now
$(b^n+c^n)^2=\big[a^nr^n(1+r^n)\big]^2=a^{2n}r^{2n}(1+r^n)^2$.....(i)
$(a^n+b^n). (c^n+d^n)=a^n(1+r^n)a^n.r^{2n}(1+r^n)$
$\qquad\qquad\:\:=a^{2n}.r^{2n}(1+r^n)^2$.......(ii)
$\therefore$ From (i) and (ii) $ (b^n+c^n)^2=(a^n+b^n). (c^n+d^n)$
$\Rightarrow\:(a^n+b^n),(b^n+c^n),(c^n+d^n)$ are in G.P.
Hence proved.