Let $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}, \frac{c_{1}}{c_{2}}$ be the consecutive terms of an arithmetic progression. If $a_{1} x^{2} + 2 b_{1} x + c_{1} = 0 and a_{2} x^{2} + 2 b_{2} x + c_{2} = 0$ have a common root, then $a_{2}, b_{2}, c_{2}$ are in

A . P .

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G . P .

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H . P .

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A . G . P .

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Solution

The correct option is B $G . P .$
Let $α$ is the common root
$a_{1} α^{2} + 2 b_{1} α + c_{1} = 0$
$a_{2} α^{2} + 2 b_{2} α + c_{2} = 0$
Solving as a simultaneous equation
$\frac{α^{2}}{2 (b_{1} c_{2} - b_{2} c_{1})} = \frac{α}{(c_{1} a_{2} - c_{2} a_{1})} = \frac{1}{2 (a_{1} b_{2} - a_{2} b_{1})}$

$α^{2} = \frac{b_{1} c_{2} - b_{2} c_{1}}{(a_{1} b_{2} - a_{2} b_{1})} \dots (1)$

$α = \frac{c_{1} a_{2} - c_{2} a_{1}}{2 (a_{1} b_{2} - a_{2} b_{1})} \dots (2)$

From $(1)$ and $(2)$

$\frac{(c_{1} a_{2} - c_{2} a_{1})^{2}}{4 (a_{1} b_{2} - a_{2} b_{1})^{2}} = \frac{b_{1} c_{2} - b_{2} c_{1}}{(a_{1} b_{2} - a_{2} b_{1})}$

$\Rightarrow (c_{1} a_{2} - c_{2} a_{1})^{2} = 4 (a_{1} b_{2} - a_{2} b_{1}) (b_{1} c_{2} - b_{2} c_{1})$

$\Rightarrow (c_{2} a_{2})^{2} \times {(\frac{c_{1} a_{2}}{c_{2} a_{2}} - \frac{c_{2} a_{1}}{c_{2} a_{2}})}^{2} = 4 \times a_{2} b_{2} (\frac{a_{1} b_{2}}{a_{2} b_{2}} - \frac{a_{2} b_{1}}{a_{2} b_{2}}) \times b_{2} c_{2} \times (\frac{b_{1} c_{2}}{b_{2} c_{2}} - \frac{b_{2} c_{1}}{b_{2} c_{2}})$

$\Rightarrow (c_{2} a_{2})^{2} \times {(\frac{c_{1}}{c_{2}} - \frac{a_{1}}{a_{2}})}^{2} = 4 \times a_{2} b_{2} (\frac{a_{1}}{a_{2}} - \frac{b_{1}}{b_{2}}) \times b_{2} c_{2} \times (\frac{b_{1}}{b_{2}} - \frac{c_{1}}{c_{2}})$

$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}, \frac{c_{1}}{c_{2}}$ are in A.P.
Let common difference be $d$

$(c_{2} a_{2})^{2} \times (2 d)^{2} = 4 a_{2} b_{2} \times (- d) \times b_{2} c_{2} \times (- d)$
$\Rightarrow c_{2} a_{2} = b_{2}^{2}$

Hence $a_{2}, b_{2}, c_{2}$ are in $G . P .$

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