Question

Given that $a{\alpha }^2+2b\beta +c\ne 0$ 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 $a,b,c$ lie in

• A. arithmetic progression
• B. geometric progression
• C. harmonic progression
• D. arithmetico-geometric progression

Answer 1

The correct option is B geometric progression

$\left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ 0& a\alpha +b& b\alpha +c\end{array}\right|=0$

$R_3\to R_3-\alpha R_1-R_2$

$\left|\begin{array}{cccc}a\alpha +b& a& b& \\ b\alpha +c& b& c& \\ -(a{\alpha }^2+2b\alpha +c)& 0& 0& \end{array}\right|=0$

$\Rightarrow -[a{\alpha }^2+2b\alpha +c]\cdot [\alpha c-b^2]=0$

$\Rightarrow ac-b^2=0$ [As $a{\alpha }^2+2b\alpha +c\ne 0]$

$\Rightarrow ac=b^2$

$\therefore$, $a,b,c$ form a geometric progression