If $a, b, c, d$ be in G.P., prove that
$(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}$

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Solution

$a, b, c, d$ are in G.P.
$\Rightarrow$ $b^{2} = a c,$ $c^{2} = b d$ and $a d = b c$
To prove:- $(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}$
L.H.S $= 2 b^{2} + 2 c^{2} + a^{2} + d^{2} - 2 b c - 2 a c - 2 b d$
Using above results
L.H.S $= 2 b^{2} + 2 c^{2} + a^{2} + d^{2} - 2 a d - 2 b^{2} - 2 c^{2}$
L.H.S $= a^{2} + d^{2} - 2 a d = (a - d)^{2} =$ R.H.S
Hence proved.

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