If $\frac{p + q}{p - q}, x$ and $p^{2} - q^{2}$ are in continued proportion then $x$ is

(p + q)

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(p - q)

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(2 p + q)

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(p + 2 q)

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Solution

The correct option is A $(p + q)$
If $a, b, c$ are in continued proportion, then $b^{2} = a c$ or $b = \sqrt{a c}$
So, $x = \sqrt{\frac{p + q}{p - q} \times (p^{2} - q^{2})}$
Formula
$a^{2} - b^{2} = (a + b) (a - b)$
Hence,
$x = \sqrt{\frac{p + q}{p - q} \times ((p + q) (p - q))}$
$x = \sqrt{(p + q) (p + q))}$
$x = p + q$

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Similar questions

\frac{p + q}{p - q}

\frac{p + q}{p - q}, x a n d p^{2} - q^{2}

P = \frac{x}{x + y}, Q = \frac{y}{x + y}

\frac{1}{P - Q} - \frac{2 Q}{P^{2} - Q^{2}}

2 [(x - p) (x - q) + (p - x) (p - q) + (q - x) (q - p)] = (p - q)^{2} + (x - p)^{2} + (x - q)^{2}

3 α 3

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

P^{2} + Q^{2} | =

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