Question

If $$a, b, c, d$$ are in continued proportion, prove that $$\left(\frac{a - b}{c} + \frac{a - c}{b}\right)^2 - \left(\frac{d - b}{c} + \frac{d - c}{b}\right)^2 = (a - d)^2 \left(\frac{1}{c^2} + \frac{1}{b^2}\right)$$.

\[ [\textbf{Hint :}\ \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \Rightarrow c = dk,\ b = dk^{2} \ \text{and} \ a = dk^{3}.\,] \]

Answer 1

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

To prove: $$\left(\frac{a - b}{c} + \frac{a - c}{b}\right)^2 - \left(\frac{d - b}{c} + \frac{d - c}{b}\right)^2 = (a - d)^2 \left(\frac{1}{c^2} + \frac{1}{b^2}\right)$$

Proof:

1. Let $$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$$, which gives $$c = dk$$, $$b = dk^2$$ and $$a = dk^3$$.
2. $$\frac{a - b}{c} = \frac{dk^3 - dk^2}{dk} = k^2 - k$$, and $$\frac{a - c}{b} = \frac{dk^3 - dk}{dk^2} = k - \frac{1}{k}$$
3. $$\frac{a - b}{c} + \frac{a - c}{b} = k^2 - \frac{1}{k}$$
4. $$\frac{d - b}{c} = \frac{d - dk^2}{dk} = \frac{1}{k} - k$$, and $$\frac{d - c}{b} = \frac{d - dk}{dk^2} = \frac{1}{k^2} - \frac{1}{k}$$
5. $$\frac{d - b}{c} + \frac{d - c}{b} = \frac{1}{k^2} - k$$
6. $$\text{L.H.S.} = \left(k^2 - \frac{1}{k}\right)^2 - \left(\frac{1}{k^2} - k\right)^2 = \left(k^4 - 2k + \frac{1}{k^2}\right) - \left(\frac{1}{k^4} - \frac{2}{k} + k^2\right) = k^4 - k^2 - \frac{1}{k^4} + \frac{1}{k^2} - 2k + \frac{2}{k}$$
7. $$\text{R.H.S.} = (dk^3 - d)^2 \left(\frac{1}{d^2 k^2} + \frac{1}{d^2 k^4}\right) = d^2(k^3 - 1)^2 \cdot \frac{1}{d^2}\left(\frac{1}{k^2} + \frac{1}{k^4}\right) = (k^3 - 1)^2 \frac{k^2 + 1}{k^4} = (k^2 - \frac{1}{k})^2 - (\frac{1}{k^2} - k)^2$$
8. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.