If $a, b, c, d$ are in continued proportion, prove that
$(a^{2} - b^{2}) (c^{2} - d^{2}) = {(b^{2} - c^{2})}^{2}$

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Solution

Given $a, b, c, d$ are in continued proportion
$⟹ \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k ($ say $)$
$⟹ c = d k, b = c k = k^{2} d, a = b k = k^{3} d$
$L H S = (a^{2} - b^{2}) (c^{2} - d^{2}) = (k^{6} d^{2} - k^{4} d^{2}) (k^{2} d^{2} - d^{2}) = k^{4} d^{4} (k^{2} - 1)^{2} = (k^{4} d^{2} - k^{2} d^{2})^{2} = ((k^{2} d)^{2} - (k d)^{2})^{2}$
$= (b^{2} - c^{2})^{2} = R H S$
Hence Proved

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