Question

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

• A. Are in A.P.
• B. Are in G. P.
• C. Are in H. P.
• D. Satisfy a +2b + 3c = 0

Answer 1

The correct option is C Are in H. P.

$\left|\begin{array}{ccc}1& 2a& a\\ 1& 3b& b\\ 1& 4c& c\end{array}\right|=0,[C_2\to C_2-2C_3]\Rightarrow \left|\begin{array}{ccc}1& 0& a\\ 1& b& b\\ 1& 2c& c\end{array}\right|=0,[R_3\to R_3-R_2,R_2\to R_2-R_1]\Rightarrow \left|\begin{array}{ccc}1& 0& a\\ 0& b& b-a\\ 0& 2c-b& c-b\end{array}\right|=0;b(c-b)-(b-a)(2c-b)=0$
On simplification,$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\therefore$ a, b, c are in Harmonic progression.