If $a, b, c$ and $d$ are in geometric sequence, then prove that $(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}$

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Solution

Let first term be $a$ and common ratio $r$ .
$b = a r, c = a r^{2}, d = a r^{3}$
L.H.S. $= (b - c)^{2} + (c - a)^{2} + (d - b)^{2}$
$= (a r - a r^{2})^{2} + (a r^{2} - a)^{2} + (a r^{3} - a r)^{2}$
$= a^{2} (r - r^{2})^{2} + a^{2} (r^{2} - 1)^{2} + a^{2} (r^{3} - r)^{2}$
$= a^{2} [(r - r^{2})^{2} + (r^{2} - 1)^{2} + (r^{3} - r)^{2}]$
$= a^{2} (r^{2} - 2 r^{3} + r^{4} + r^{4} - 2 r^{3} + 1 + r^{6} - 2 r^{4} + r^{2})$
$= a^{2} (r^{6} - 2 r^{3} + 1)$
$= a^{2} (1 - r^{3})^{2}$ .... (1)
$(a - d)^{2} = (a - a r^{3})^{2}$
$= a^{2} (1 - r^{3})^{2}$ ....(2)
From (1) and (2), we have
$(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}$

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