Question
If the system of linear equations $$x + 2ay + az = 0 ; x + 3by + bz = 0 ; x + 4cy + cz = 0 ;$$ has a non - zero solution, then $$a, b, c.$$
A.
satisfy $$a + 2b + 3c = 0$$
B.
are in A.P.
C.
are in G.P.
D.
are in H.P.
Answer :
are in H.P.
Solution :
For homogeneous system of equations to have non-zero solution, $$\Delta = 0$$
\[\begin{array}{l}
\left| \begin{array}{l}
1\,\,\,\,\,\,2a\,\,\,\,\,\,a\\
1\,\,\,\,\,\,3b\,\,\,\,\,\,\,b\\
1\,\,\,\,\,\,4c\,\,\,\,\,\,c
\end{array} \right| = 0\,\,\,{C_2} \to {C_2} - 2{C_3}\\
\left| \begin{array}{l}
1\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,a\\
1\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,b\\
1\,\,\,\,\,\,2c\,\,\,\,\,\,\,c
\end{array} \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}\\
\left| \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b - a\\
0\,\,\,\,\,\,2c\, - b\,\,\,\,\,\,\,\,c - b
\end{array} \right| = 0
\end{array}\]
$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$
On simplification, $$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
∴ $$a, b, c$$ are in Harmonic Progression.