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If a, b, c are in continued proportion, prove that: (iii) \[\mathbf{a}:\text{ }\mathbf{c}\text{ }=\text{ }({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}):\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\] (iv) \[~{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}~({{\mathbf{a}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{-\mathbf{4}}})\text{ }=\text{ }{{\mathbf{b}}^{-\mathbf{2}}}~({{\mathbf{a}}^{\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{\mathbf{4}}})\]
If a, b, c are in continued proportion, prove that: (iii) \[\mathbf{a}:\text{ }\mathbf{c}\text{ }=\text{ }({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}):\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\] (iv) \[~{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}~({{\mathbf{a}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{-\mathbf{4}}})\text{ }=\text{ }{{\mathbf{b}}^{-\mathbf{2}}}~({{\mathbf{a}}^{\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{\mathbf{4}}})\]
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If a, b, c are in continued proportion, prove that: (iii) \[\mathbf{a}:\text{ }\mathbf{c}\text{ }=\text{ }({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}):\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\] (iv) \[~{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}~({{\mathbf{a}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{-\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{-\mathbf{4}}})\text{ }=\text{ }{{\mathbf{b}}^{-\mathbf{2}}}~({{\mathbf{a}}^{\mathbf{4}}}~+\text{ }{{\mathbf{b}}^{\mathbf{4}}}~+\text{ }{{\mathbf{c}}^{\mathbf{4}}})\]

It is given that

a, b, c are in continued proportion

So we get

a/b = b/c = k

 

(iii) \[~a:\text{ }c\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}):\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\]

We can write it as

Therefore, LHS = RHS.

(iv) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}~({{a}^{-4}}~+\text{ }{{b}^{-4}}~+\text{ }{{c}^{-4}})\text{ }=\text{ }{{b}^{-2}}~({{a}^{4}}~+\text{ }{{b}^{4}}~+\text{ }{{c}^{4}})\]

Therefore, LHS = RHS.

i

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