It is given that $a, b, c$ and $d$ are in $G.P.$

$\therefore b^{2}=a c$       ……..$(1)$

$c^{2}=b d$       ……..$(2)$

$a d=b c$       ……..$(3)$

It has to be proved that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)$ are in $G.P.$ i.e.,

$\left(b^{n}+c^{n}\right)^{2}=\left(a^{n}+b^{n}\right),\left(c^{n}+d^{n}\right)$

Consider $L.H.S.$

$\left(b^{n}+c^{n}\right)^{2}=b^{2 n}+2 b^{n} c^{n}+c^{2 n}$

$=\left(b^{2}\right)^{n}+2 b^{n} c^{n}+\left(c^{2}\right)^{n}$

$=(a c)^{n}+2 b^{n} c^{n}+(b d)^{n}$            [ 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 $(3)$ ]

$=c^{n}\left(a^{n}+b^{n}\right)+d^{n}\left(a^{n}+b^{n}\right)$

$=\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)=$ $\mathrm{R.H.S.}$

$\therefore\left(b^{n}+c^{n}\right)^{2}=\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)$

Thus, $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),$ and $\left(c^{n}+d^{n}\right)$ are in $G.P.$