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If a, b, c, d are in continued proportion, prove that: (V) \[{{\left( \frac{a-b}{c}+\frac{a-c}{b} \right)}^{2}}-{{\left( \frac{d-b}{c}+\frac{d-c}{b} \right)}^{2}}={{(a-d)}^{2}}\left( \frac{1}{{{c}^{2}}}-\frac{1}{{{b}^{2}}} \right)\]
If a, b, c, d are in continued proportion, prove that: (V) \[{{\left( \frac{a-b}{c}+\frac{a-c}{b} \right)}^{2}}-{{\left( \frac{d-b}{c}+\frac{d-c}{b} \right)}^{2}}={{(a-d)}^{2}}\left( \frac{1}{{{c}^{2}}}-\frac{1}{{{b}^{2}}} \right)\]
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If a, b, c, d are in continued proportion, prove that: (V) \[{{\left( \frac{a-b}{c}+\frac{a-c}{b} \right)}^{2}}-{{\left( \frac{d-b}{c}+\frac{d-c}{b} \right)}^{2}}={{(a-d)}^{2}}\left( \frac{1}{{{c}^{2}}}-\frac{1}{{{b}^{2}}} \right)\]

It is given that

a, b, c, d are in continued proportion

Here we get

a/b = b/c = c/d = k

\[c\text{ }=\text{ }dk,\text{ }b\text{ }=\text{ }ck\text{ }=\text{ }dk\text{ }.\text{ }k\text{ }=\text{ }d{{k}^{2}}\]

\[a\text{ }=\text{ }bk\text{ }=\text{ }d{{k}^{2}}~.\text{ }k\text{ }=\text{ }d{{k}^{3}}\]

Therefore, LHS = RHS.

i

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