State true or false.
If $a, b$ and $c$ are in continued proportion, then $a : c = (a^{2} + b^{2}) : (b^{2} + c^{2})$

True

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False

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Solution

The correct option is A True
Given,

a, b and c are the continued proportion

$a : b = b : c$

$\Rightarrow \frac{a}{b} = \frac{b}{c}$

$\Rightarrow b^{2} = a c$

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

$\Rightarrow a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

L.H.S. $= a (b^{2} + c^{2}) \Rightarrow a (a c + c^{2}) \Rightarrow a c (a + c)$

R.H.S. $= c (a^{2} + b^{2}) \Rightarrow c (a^{2} + a c) \Rightarrow a c (a + c)$

L.H.S. = R.H.S. Hence proved.

Suggest Corrections

Similar questions

a, b, c

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

\frac{a^{2} + b^{2} + c^{2}}{{(a + b + c)}^{2}} = \frac{a - b + c}{a + b + c}

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

If a, b, and c are in continued proportion, then $a \times c = b^{2}$ .

Continued proportion means a:b :: b:c

If a + b > c, then, $a^{2} + b^{2} > c^{2}$ for any numbers a,b,c. State true or false.

a, b, c, d

a - c : b - d = \sqrt{a^{2} + c^{2}} : \sqrt{b^{2} + d^{2}}

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