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If the system of linear equations $x + 2ay + az = 0$ $x + 3by + bz = 0$ $x + 4cy + cz = 0$ has a non-zero solution, then $a, b, c$
are in $G..P.$
are in $H.P.$
satisfy $a + 2b + 3c = 0$
are in $A.P.$
Solution
The system of linear equations has a non-zero solution, then $\left|\begin{array}{lll}1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c\end{array}\right|=0$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$\begin{array}{ccc}1 & 2 a & a \\ 0 & 3 b-2 a & b-a \\ 0 & 4 c-2 a & c-a\end{array} \mid=0$
$\Rightarrow(3 b-2 a)(c-a)-(4 c-2 a)(b-a)=0$
$\Rightarrow 3 b c-3 b a-2 a c+2 a^{2}=4 b c-2 a b-4 a c+2 a^{2}$
$\Rightarrow 2 a c=b c+a b$
On dividing by abc, we get $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
Hence, a, b, c are in HP.
