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

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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 $)$
$⟹ c = d k, b = c k = k^{2} d, a = b k = k^{3} d$
$L H S = (a + d) (b + c) - (a + c) (b + d) = (k^{3} d + d) (k^{2} d + k d) - (k^{3} d + k d) (k^{2} d + d)$
$= d^{2} (k^{5} + k^{2} + k^{4} + k - k^{5} - 2 k^{3} - k)$
$= k^{2} d^{2} (k - 1)^{2} = (k^{2} d - k d)^{2} = (b - c)^{2} = R H S$
Hence Proved

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