Question

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

Answer 1

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

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

Proof:

1. Since $$a, b, c$$ are in continued proportion, let $$\frac{a}{b} = \frac{b}{c} = k$$, so $$b = ck$$ and $$a = ck^2$$.
2. $$\text{L.H.S.} = \frac{ck^2 + ck + c}{ck^2 - ck + c} = \frac{c(k^2 + k + 1)}{c(k^2 - k + 1)} = \frac{k^2 + k + 1}{k^2 - k + 1}$$
3. $$\text{R.H.S.} = \frac{(ck^2 + ck + c)^2}{(ck^2)^2 + (ck)^2 + c^2} = \frac{c^2(k^2 + k + 1)^2}{c^2(k^4 + k^2 + 1)}$$
4. $$\text{R.H.S.} = \frac{(k^2 + k + 1)^2}{(k^2 + k + 1)(k^2 - k + 1)} = \frac{k^2 + k + 1}{k^2 - k + 1}$$ [$$\because k^4 + k^2 + 1 = (k^2 + k + 1)(k^2 - k + 1)$$]
5. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.