If {a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac) , where a,b,c are non-zero numbers. Then a,b,c are

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8. Sequences and Series

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If ${a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac)$, where $a,b,c$ are non-zero numbers. Then $a,b,c$ are in

A

$A.P$

B

$G.P$

C

$H.P$

D

None of these

Solution

(b) $(a + 2b + 2c)$$(a – 2b + 2c) = {a^2} + 4{c^2}$

$⇒$ ${(a + 2c)^2} – {(2b)^2} = {a^2} + 4{c^2}$

$⇒$ ${a^2} + 4ac + 4{c^2} – 4{b^2} = {a^2} + 4{c^2}$

$⇒$ $4ac – 4{b^2} = 0$

$⇒$ ${b^2} = ac$

Hence $a, b, c$ are in $G.P.$

Standard 11

Mathematics

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