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

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Solution

Given $a, b, c, d$ are in a geometric sequence.
Let $r$ be the common ratio of the given sequence.
Here, the first term is $a$ .
Thus, $b = a r, c = a r^{2}, d = a r^{3}$
Now, $(b - c)^{2} + (c - a)^{2} + (d - b)^{2}$
$= (a r - a r^{2})^{2} + (a r^{2} - a)^{2} + (a r^{3} - a r)^{2}$
$= a^{2} [(r - r^{2})^{2} + (r^{2} - 1)^{2} + (r^{3} - r)^{2}]$
$= a^{2} [r^{2} - 2 r^{3} + r^{4} + r^{4} - 2 r^{2} + 1 + r^{6} - 2 r^{4} + r^{2}]$
$= a^{2} [r^{6} - 2 r^{3} + 1] = a^{2} [r^{3} - 1]^{2}$
$= (a r^{3} - a)^{2} = (a - a r^{3})^{2} = (a - d)^{2}$

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