If p,q and r are in A.P, then which of the following is correct?

p q + q r - 2 p^{2} = 0

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p q + q r - 2 q^{2} = 0

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p q + q r - 2 r^{2} = 0

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p q + q r - 2 p q r = 0

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Solution

The correct option is B $p q + q r - 2 q^{2} = 0$
Since p,q and r are in A.P
$p + r = 2 q$
Multiplying $q$ on both sides
$\Rightarrow q (p + r) = 2 q^{2}$ .
$∴ p q + q r - 2 q^{2} = 0$

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Similar questions

⎡ ⎢ ⎣ \begin{matrix} - q r & p (q + r) & p r + p q p q + q r & - p r & p q + q r q r + p r & q r + p r & - p q \end{matrix} ⎤ ⎥ ⎦

| A |

(p q - q r + p r) (p q + q r) - (p r + p q) (p + q - r)

p, q, r

p + q + r = 0, a (q r + r p + p q) = 2 a - x, a p q r = y, q r = 1

Δ P Q R

∠ R > ∠ Q

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