If $a, b, c, d$ are in $G.P.$,prove that $(a^{n}+b^{n}), (b^{n}+c^{n}), (c^{n}+d^{n})$ are in $G.P.$

(N/A) It is given that $a, b, c,$ and $d$ are in $G.P.$
$\therefore b^{2}=ac$ ........$(1)$
$c^{2}=bd$ ........$(2)$
$ad=bc$ ........$(3)$
It has to be proved that $(a^{n}+b^{n}), (b^{n}+c^{n}), (c^{n}+d^{n})$ are in $G.P.$,i.e.,
$(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Consider $L.H.S.$
$(b^{n}+c^{n})^{2}=b^{2n}+2b^{n}c^{n}+c^{2n}$
$=(b^{2})^{n}+2b^{n}c^{n}+(c^{2})^{n}$
$=(ac)^{n}+2b^{n}c^{n}+(bd)^{n}$ [ Using $(1)$ and $(2)$ ]
$=a^{n}c^{n}+b^{n}c^{n}+b^{n}c^{n}+b^{n}d^{n}$
$=a^{n}c^{n}+b^{n}c^{n}+a^{n}d^{n}+b^{n}d^{n}$ [ Using $(3)$,since $bc=ad$,so $b^{n}c^{n}=a^{n}d^{n}$ ]
$=c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$
$=(a^{n}+b^{n})(c^{n}+d^{n}) = R.H.S.$
$\therefore (b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Thus,$(a^{n}+b^{n}), (b^{n}+c^{n}),$ and $(c^{n}+d^{n})$ are in $G.P.$