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

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Solution

To prove ${(b - c)}^{2} + {(c - a)}^{2} + {(d - b)}^{2} = {(a - d)}^{2}$
Since $a, b, c, d$ are in G.P.
$\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r$
$c = b r; b^{2} = c a \Rightarrow c = \frac{b^{2}}{a}$
$b = a r$
$\frac{b}{a} = \frac{d}{c}$
$b c = a d$
$\frac{b \times b^{2}}{a \times a} = d$
$d = \frac{b^{3}}{a^{2}}$

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