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
Answer & explanation
Correct answer: option 4
The correct answer is option (4) : are in H.P.
The given system of equations will have a non-zero solution, if
$\begin{vmatrix}1 & 2a & 3\\1 & 3b & b\\1 & 4c & c\end{vmatrix}=0$
$⇒\begin{vmatrix}1 & 2a & a\\0 & 3b-2a & b-a\\0 & 4c-2a & c-a\end{vmatrix}$ Applying $R_2→R_2-R_1\, R_3→R_3-R_1$
$⇒(3b-2a) (c-a) -(b-a) (4c-2a) = 0 $
$⇒(3bc-3ab-2ca+2a^2) - ( 4bc - 2ab - 4ca +2a^2 ) = 0 $
$⇒ -bc - ab + 2 ca = 0 $
$⇒ \frac{2}{b}=\frac{1}{a}+\frac{1}{c}⇒a,b,c $ are in H.P.