Question

Given that, $a{\alpha }^2+2b\alpha +c\ne 0$ and that the
system of equations
$(a\alpha +b)x+ay+bz=0;$
$(b\alpha +c)x+by+cz=0;$
$(a\alpha +b)y+(b\alpha +c)z=0$
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
$(a\alpha +b)x+ay+bz=0$
$(b\alpha +c)x+by+cz=0$
and $(a\alpha +b)y+(b\alpha +c)z=0$
For non-trivial solution,
$\left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ 0& a\alpha +b& b\alpha +c\end{array}\right|=0$
Applying $R_3\to R_3-\alpha R_1-R_2$
$\Rightarrow \left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ -\left(a{\alpha }^2+2b\alpha +c\right)& 0& 0\end{array}\right|=0$
$\Rightarrow -\left(a{\alpha }^2+2b\alpha +c\right)\left(ac-b^2\right)=0$
$\Rightarrow ac-b^2=0(\because a{\alpha }^2+2b\alpha +c\ne 0)$
$\Rightarrow ac=b^2$
Hence, $a,b$ and $c$ are in GP.