Q.If \(a, b, c, d\) are in continuous proportion, prove that \((a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2\).

Assume, \(\cfrac{a}{b} = \cfrac{b}{c} = \cfrac{c}{d} = k\) [\(k\) is a nonzero constant] \(\therefore a = bk, \quad b = ck, \quad c = dk\) \(\therefore b = dk \times k = dk^2, \quad a = dk^2 \times k = dk^3\) ### Left-hand side: \((a^2 - b^2)(c^2 - d^2)\) \(= \{(dk^3)^2 - (dk^2)^2\} \{(dk)^2 - d^2\}\) \(= \{d^2k^6 - d^2k^4\} \{d^2k^2 - d^2\}\) \(= d^2k^4(k^2 - 1) \times d^2(k^2 - 1)\) \(= d^4k^4(k^2 - 1)^2\) ### Right-hand side: \((b^2 - c^2)^2\) \(= \{(dk^2)^2 - (dk)^2\}^2\) \(= \{d^2k^4 - d^2k^2\}^2\) \(= \{d^2k^2(k^2 - 1)\}^2\) \(= d^4k^4(k^2 - 1)^2\) \(\therefore\) Left-hand side = Right-hand side [Proved]
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