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Mathematics

If a, b and c are in continued proportion, then prove that :

3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}.

Ratio Proportion

ICSE Sp 2024

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Answer

Since, a, b and c are in continued proportion.

ab=bcb2=ac.\therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow b^2 = ac.

Substituting, b2 = ac in L.H.S. of equation 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}, we get :

3a2+5ab+7ac3ac+5bc+7c2a(3a+5b+7c)c(3a+5b+7c)ac.\Rightarrow \dfrac{3a^2 + 5ab + 7ac}{3ac + 5bc + 7c^2} \\[1em] \Rightarrow \dfrac{a(3a + 5b + 7c)}{c(3a + 5b + 7c)} \\[1em] \Rightarrow \dfrac{a}{c}.

Hence, proved that 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}.

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