Question

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

• A. Arithmetic progression
• B. Geometric progression
• C. Harmonic progression
• D. Arithmetico-geometric progression

Answer 1

Answer: (B)

Given system of equations is
$(a \alpha+b) x+a y+b z=0$
$ (b \alpha+c) x+b y+c z=0$
and $(a \alpha+b) y+(b \alpha+c) z=0$
For non-trivial solution,
$\begin{vmatrix} a \alpha+b & a & b \\ b \alpha+c & b & c \\ 0 & a \alpha+b & b \alpha+c \end{vmatrix} =0$
Applying $R_{3} \rightarrow R_{3}-\alpha R_{1}-R_{2}$
$\Rightarrow \begin{vmatrix}a \alpha+b & a & b \\ b \alpha+c & b & c \\ -\left(a \alpha^{2}+2 b \alpha+c\right) & 0 & 0\end{vmatrix} =0$
$\Rightarrow -\left(a \alpha^{2}+2 b \alpha+c\right)\left(a c-b^{2}\right)=0$
$\Rightarrow a c-b^{2}=0 (\because a \alpha^{2}+2 b \alpha+c \neq 0)$
$\Rightarrow a c=b^{2}$
Hence, $a, b$ and $c$ are in GP.