Question

If the system of linear equations $x+2ay+az=0,x+3by+bz=0,$ and
$x+4cy+cz=0$ has a non-zero solution, then a, b, c

• A. Are in AP
• B. Are in GP
• C. Are in HP
• D. Satisfy $a+2b+3c=0$

Answer 1

Answer: (C)

Are in HP

The system of linear equations has a non-zero solution, then
$\left|\begin{array}{ccc}1& 2a& a\\ 1& 3b& b\\ 1& 4c& c\end{array}\right|=0$
Applying R${}_2\to$R${}_2$ - R${}_1$, R${}_3\to$ R${}_3$ - R${}_1$
$\left|\begin{array}{ccc}1& 2a& a\\ 0& 3b-2a& b-a\\ 0& 4c-2a& c-a\end{array}\right|=0$
$\Rightarrow (3b-2a)(c-a)-(4c-2a)(b-a)=0$
$\Rightarrow 3bc-3ba-2ac+2a^2=4bc-2ab-4ac+2a^2$
$\Rightarrow 2ac=bc+ab$
On dividing by abc, we get
$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
Hence, a, b, c are in HP.