If a- b- c- d are nonzero real numbers such that a2-b2-c2b2-c2-d2- ab-bc-cd2- then a- b- c- d are in-

The correct option is C GP (a^2+b^2+c^2)(b^2+c^2+d^2)≤ (ab+bc+cd)^2 Solving this we get b^4+c^4+a^2c^2+a^2b^2+b^2d^2+b^2c^2 2ab^2c 2bc^2d 2abcd ≤ 0 or ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If a, b, c,...

Question

If $a, b, c, d$ are nonzero real numbers such that $(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) \leq (a b + b c + c d)^{2}$ , then $a, b, c, d$ are in-

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Solution

The correct option is C GP
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) \leq (a b + b c + c d)^{2}$
Solving this we get
$b^{4} + c^{4} + a^{2} c^{2} + a^{2} b^{2} + b^{2} d^{2} + b^{2} c^{2} - 2 a b^{2} c - 2 b c^{2} d - 2 a b c d \leq 0$
or $(b^{2} - a c)^{2} + (c^{2} - d b)^{2} + (a d - b c)^{2} \leq 0$
$∴ b^{2} - a c = 0 \Rightarrow \frac{b}{a} = \frac{c}{b}, c^{2} - b d = 0 \Rightarrow \frac{c}{b} = \frac{d}{c}, a b - b c = 0 \Rightarrow \frac{d}{c} = \frac{b}{a}$
$∴ \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$ Hence $a, b, c, d$ are in G.P

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Similar questions

Q. If a, b, c, d are non zero real numbers such that

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

then a, b, c, d are in

If a, b, c, d are in GP, prove that

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

Q. If a, b, c and d are in G.P. show that

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

a, b, c, d

are in continued proportion, prove that

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

Q. If

a, b, c, d, x

are distinct real numbers such that

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

then

a, b, c, d

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If a,b,c,d are nonzero real numbers such that (a2+b2+c2)(b2+c2+d2)≤(ab+bc+cd)2, then a,b,c,d are in-

The correct option is C GP(a2+b2+c2)(b2+c2+d2)≤(ab+bc+cd)2Solving this we getb4+c4+a2c2+a2b2+b2d2+b2c2−2ab2c−2bc2d−2abcd≤0or (b2−ac)2+(c2−db)2+(ad−bc)2≤0∴b2−ac=0⇒ba=cb,c2−bd=0⇒cb=dc,ab−bc=0⇒dc=ba∴ba=cb=dc Hence a,b,c,d are in G.P

If $a, b, c, d$ are nonzero real numbers such that $(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) \leq (a b + b c + c d)^{2}$ , then $a, b, c, d$ are in-