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Question
If $$a, b, c$$ are in continued proportion, prove that $$\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c}$$.
Theorem
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Solution
Given: $$a, b, c$$ are in continued proportion.
To prove: $$\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c}$$
Proof:
- Let $$\frac{a}{b} = \frac{b}{c} = k$$, then $$b = ck$$ and $$a = ck^2$$.
- $$\text{L.H.S.} = \frac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} = \frac{c^2 k^4 + c^2 k^3 + c^2 k^2}{c^2 k^2 + c^2 k + c^2}$$
- $$\text{L.H.S.} = \frac{c^2 k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} = k^2$$
- $$\text{R.H.S.} = \frac{a}{c} = \frac{ck^2}{c} = k^2$$
- $$\text{L.H.S.} = \text{R.H.S.}$$
Hence proved.
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Chapter 7: Ratio and Proportion - EXERCISE 7B [Page 104]
