Question

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

• 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

Answer: (C)

To have non-trivial solution, $\begin{vmatrix}1&4a&a\\ 1&3b&b\\ 1&2c&c\end{vmatrix} = 0$
$ \Rightarrow 1(3bc - 2bc) - (4ac - 2ac) + 1 (4ab - 3ab) = 0$
$\Rightarrow bc - 2ac + ab = 0$
$\Rightarrow bc + ab = 2ac$
$\Rightarrow b(a + c) = 2ac$
$\Rightarrow b = \frac{2ac}{a + c}$
$\Rightarrow a, b, c$ are in $HP$