Question

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

Answer 1

Given: $$a, b, c, d$$ are in continued proportion.

To prove: $$(a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2$$

Proof:

1. Let $$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$$, so $$c = dk$$, $$b = dk^2$$, $$a = dk^3$$.
2. $$\text{L.H.S.} = (a^2 - b^2)(c^2 - d^2) = [(dk^3)^2 - (dk^2)^2][(dk)^2 - d^2] = (d^2 k^6 - d^2 k^4)(d^2 k^2 - d^2)$$
3. $$\text{L.H.S.} = d^2 k^4(k^2 - 1) \cdot d^2(k^2 - 1) = d^4 k^4 (k^2 - 1)^2$$
4. $$\text{R.H.S.} = (b^2 - c^2)^2 = [(dk^2)^2 - (dk)^2]^2 = (d^2 k^4 - d^2 k^2)^2 = [d^2 k^2(k^2 - 1)]^2 = d^4 k^4 (k^2 - 1)^2$$
5. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.