If a- b- c- d and p are different real numbers such that-a2 - b2 - c2- p2 minus- 2 -ab - bc - cd- p - -b2 - c2 - d2- le- 0- then show that a- b- c and d are in G-P-

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Byju's Answer

Standard XI

Economics

Relationship between the Short Run Average and Marginal Cost Curves

If a, b, c, d...

Question

If a, b, c, d and p are different real numbers such that:
(a² + b² + c²) p² − 2 (ab + bc + cd) p + (b² + c² + d²) ≤ 0, then show that a, b, c and d are in G.P.

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Solution

$(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0 \Rightarrow (a^{2} p^{2} + b^{2} p^{2} + c^{2} p^{2}) - 2 (a b p + b c p + c d p) + (b^{2} + c^{2} + d^{2}) \leq 0 \Rightarrow (a^{2} p^{2} - 2 a b p + b^{2}) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) \leq 0 \Rightarrow {(a p - b)}^{2} + {(b p - c)}^{2} + {(c p - d)}^{2} \leq 0 \Rightarrow {(a p - b)}^{2} + {(b p - c)}^{2} + {(c p - d)}^{2} = 0 \Rightarrow {(a p - b)}^{2} = 0 \Rightarrow p = \frac{b}{a} Also, {(b p - c)}^{2} = 0 \Rightarrow p = \frac{c}{b} Similiarly, \Rightarrow {(c p - d)}^{2} = 0 \Rightarrow p = \frac{d}{c} ∴ \frac{b}{a} = \frac{c}{b} = \frac{d}{c} Thus, a, b, c and d are in G . P .$

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Similar questions

Q. If a, b, c, d and p are different real numbers such that:
(a² + b² + c²) p² − 2 (ab + bc + cd) p + (b² + c² + d²) ≤ 0, then show that a, b, c and d are in G.P.

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

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

are in G.P.

Q. If

a, b, c, d

and

p

are distinct real numbers 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

G . P .

Q. If

a, b, c, d

and

p

are distinct real numbers 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

and

d

Q. 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 + d c) p + (b^{2} + c^{2} + d^{2}) \geq 0

, then

a, b, c, d

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If a, b, c, d and p are different real numbers such that: (a2 + b2 + c2) p2 − 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d are in G.P.

a2+b2+c2p2-2ab+bc+cdp+b2+c2+d2≤0⇒a2p2+b2p2+c2p2-2abp+bcp+cdp+b2+c2+d2≤0⇒a2p2-2abp+b2+b2p2-2bcp+c2+c2p2-2cdp+d2≤0⇒ap-b2+bp-c2+cp-d2≤0⇒ap-b2+bp-c2+cp-d2=0⇒ap-b2=0 ⇒p = baAlso, bp-c2=0 ⇒p = cbSimiliarly, ⇒cp-d2=0 ⇒p = dc∴ ba= cb=dcThus, a, b, c and d are in G.P.

If a, b, c, d and p are different real numbers such that:
(a² + b² + c²) p² − 2 (ab + bc + cd) p + (b² + c² + d²) ≤ 0, then show that a, b, c and d are in G.P.