Question

If the following system of linear equations have a non-trivial solution
$x+4ay+az=0$
$x+3by+bz=0$ and
$x+2cy+cz=0$, then

• A. $a,b,c$ are in $A.P.$
• B. $a,b,c$ are in $G.P.$
• C. $a,b,c$ are in $H.P.$
• D. $a+b+c=0$

Answer 1

The correct option is C $a,b,c$ are in $H.P.$

For non-trivial solution :
$\Delta =0$
$\Rightarrow \Delta =\left|\begin{array}{ccc}1& 4a& a\\ 1& 3b& b\\ 1& 2c& c\end{array}\right|$
Expanding along $C_1$, we get
$\Rightarrow \Delta =1(3bc-2bc)-1(4ac-2ac)+(4ab-3ab)=0$
$\Rightarrow bc-2ac+ab=0$
$\Rightarrow bc+ab=2ac$
$\Rightarrow b=\frac{2ac}{a+c}$
Hence $a,b,c$ are in $H.P.$