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