If a,b,c,d and p are distinct real number such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ , then a,b,c,d

are in A.P

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are in G.P

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are in H.P

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satisfy ab=cd

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satisfy none of these

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Solution

The correct option is B are in G.P
$(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$
$\Rightarrow (a^{2} p^{2} - 2 a b p + b^{2}) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) \leq 0$
$\Rightarrow (a p - b)^{2} + (b p - c)^{2} + (c p - d)^{2} \leq 0$
Sum of square for real numbers can't be negative
Thus $a p = b, b p = c, c p = d$
$\Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$
$\Rightarrow a, b, c, d ϵ$ G.P

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Similar questions

(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0

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$(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ , then show that a, b, c and d are in G.P.

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(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0.

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(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + d c) p + (b^{2} + c^{2} + d^{2}) \geq 0

a, b, c, d

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