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Question

If the sum of the roots of the equation $$ax^{2} + bx + c = 0$$ is equal to sum of the squares of their reciprocals, then $$bc^{2}, ca^{2}, ab^{2}$$ are in


A
A.P
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B
G.P
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C
H.P
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D
A.G.P
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Solution

The correct option is B $$A.P$$
Let $$\alpha$$  and $$\beta$$  are the roots of $$a{ x }^{ 2 }+bx+c=0$$
We know $$\alpha+\beta = \dfrac { -b }{ a }$$ $$,\alpha .\beta =\dfrac { c }{ a }$$ 
Given, $$\alpha +\beta =\dfrac { 1 }{ { \alpha  }^{ 2 } } +\dfrac { 1 }{ { \beta  }^{ 2 } } \quad $$
$$=\dfrac { { \alpha  }^{ 2 }+{ \beta  }^{ 2 } }{ { \alpha  }^{ 2 }.{ \beta  }^{ 2 } } $$
$$=\dfrac { { \left( \alpha +\beta  \right)  }^{ 2 }-2\alpha .\beta  }{ { \alpha  }^{ 2 }{ \beta  }^{ 2 } } $$
Put values of $$\alpha$$  and $$\beta$$, we get
$$\dfrac { -b }{ a } =\dfrac { { \left( \dfrac { -b }{ a }  \right)  }^{ 2 }-2\dfrac { c }{ a }  }{ { \left( \dfrac { { c } }{ { a } }  \right)  }^{ 2 } } $$
$$=\dfrac { \dfrac { { b }^{ 2 } }{ { a }^{ 2 } } -2a\dfrac { c }{ { a }^{ 2 } }  }{ \dfrac { { c }^{ 2 } }{ { a }^{ 2 } }  } $$
$$=\dfrac { { b }^{ 2 }-2ac }{ { c }^{ 2 } } $$
$$\Rightarrow -b{ c }^{ 2 }=a{ b }^{ 2 }-2{ a }^{ 2 }c$$
$$\Rightarrow 2{ a }^{ 2 }c=a{ b }^{ 2 }+b{ c }^{ 2 }$$
Therefore, $$ a{ b }^{ 2 },{ a }^{ 2 }c,b{ c }^{ 2 }$$ are in $$A.P.$$
Option A is correct.

Mathematics

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