If a, b, c, d are in G.P, prove that are in G.P.

Solution:

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

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

$c^{2}=b d \ldots(2)$

$a d=b c \ldots(3)$

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)]

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

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

$=$ R.H.S.

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

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