If a - b and c are in continued proportion- then a - c -A- a2-a b-b2-b2-b c-c2b2-b c-c2B- 2 a2-a b-b2-2 b2-b c-c2b2-b c-c22 b2-b c-c2C- b2-b c-c2-a2-a b-b2D- b2-a b-2 c2-a2-a b-2 b2

The correct option is A a^2+ab+b^2b^2+bc+c^2If a, b and c are in continued proportion, ∴ab=bc let, ab=bc=k ∴ b=ck, a=ck^2 LHS = ac=ck^2c=k^2 Let's consider ...

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Standard X

Mathematics

Proportion

If a , b and ...

Question

If $a, b$ and $c$ are in continued proportion, then $\frac{a}{c} =$ _____.

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}}

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\frac{2 a^{2} + a b + b^{2}}{2 b^{2} + b c + c^{2}}

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\frac{b^{2} + b c + c^{2}}{a^{2} + a b + b^{2}}

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\frac{b^{2} + b c + 2 c^{2}}{a^{2} + a b + 2 b^{2}}

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Solution

The correct option is A $\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}}$
If $a, b$ and $c$ are in continued proportion,

$∴ \frac{a}{b} = \frac{b}{c}$

let, $\frac{a}{b} = \frac{b}{c} = k$

$∴ b = c k, a = c k^{2}$
$LHS = \frac{a}{c} = \frac{c k^{2}}{c} = k^{2}$

Let's consider,
$RHS = \frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}}$

$= \frac{(k^{2} c)^{2} + k^{2} c (c k) + (c k)^{2}}{(c k)^{2} + (c k) (c) + c^{2}}$

$= \frac{k^{4} c^{2} + k^{3} c^{2} + c^{2} k^{2}}{c^{2} k^{2} + c^{2} k + c^{2}}$

$= \frac{c^{2} k^{2} (k^{2} + k + 1)}{c^{2} (k^{2} + k + 1)}$

$= k^{2}$

$∴ LHS = RHS$

$∴ \frac{a}{c} = \frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}}$

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Similar questions

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(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a b + b c + c d)^{2}

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a² + b² + c² - ab - bc - ca equals:

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(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a b + b c + c d)^{2}

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If a,b and c are in continued proportion, then ac= _____.

The correct option is A a2+ab+b2b2+bc+c2If a,b and c are in continued proportion, ∴ab=bc let, ab=bc=k ∴b=ck,a=ck2 LHS=ac=ck2c=k2 Let's consider, RHS=a2+ab+b2b2+bc+c2 =(k2c)2+k2c(ck)+(ck)2(ck)2+(ck)(c)+c2 =k4c2+k3c2+c2k2c2k2+c2k+c2 =c2k2(k2+k+1)c2(k2+k+1) =k2 ∴LHS=RHS ∴ac=a2+ab+b2b2+bc+c2

If $a, b$ and $c$ are in continued proportion, then $\frac{a}{c} =$ _____.