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

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Solution

$R . H . S = (b - c)^{2} + (c - a)^{2} + (d - b)^{2}$
Being $a, b, c$ and $d$ in $G P$ ,
$\frac{a}{b} = \frac{b}{c} = \frac{c}{d},$

$b^{2} = a c, c^{2} = b d$ and $a c = b d, a d = b c,$

$R H S = b^{2} + c^{2} - 2 b c + c^{2} + a^{2} - 2 a c + d^{2} + b^{2} - 2 b d$
$= 2 b^{2} - 2 b c + 2 c^{2} + a^{2} - 2 b^{2} + d^{2} - 2 c^{2}$
$= a^{2} + d^{2} - 2 a d$
$= (a - d)^{2}$
$= L H S$

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