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Q.

Given that $a α^{2} + 2 b α + c \neq 0$ and that the system of equations $(a α + b) x + a y + b z = 0, (b α + c) x + b y + c z = 0, (a α + b) y + (b α + c) z = 0$ has non- trivial solution then a,b,c lies in

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a

A.P

b

G.P

c

H.P

d

A.G.P

answer is B.

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Detailed Solution

$\begin{array}{l} |\begin{array}{lcc} a α + b & a & b \\ b α + c & b & c \\ 0 & a α + b & b α + c \end{array}| = 0 C_{1} \to C_{1} - α C_{2} - C_{3} \\ \Rightarrow |\begin{matrix} 0 & a & b \\ 0 & b & c \\ - (a α^{2} + 2 b α + c) & a α + b & b α + c \end{matrix}| = 0 \\ - a [0 + c (a α^{2} + 2 b α + c)] + b [0 + b (a α^{2} + 2 b α + c)] = 0 \end{array}$

$a α^{2} + 2 b α + c \neq 0$

then $b^{2} = a c$

A,b,c are in G.P

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