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If the system of linear equations
$x+2 a y+a z=0$ $x+3 b y+b z=0$
$x+4 c y+c z=0$
has a non-zero solution, then $\mathrm{a}, \mathrm{b}, \mathrm{c}$
Options:
$x+2 a y+a z=0$ $x+3 b y+b z=0$
$x+4 c y+c z=0$
has a non-zero solution, then $\mathrm{a}, \mathrm{b}, \mathrm{c}$
Solution:
1703 Upvotes
Verified Answer
The correct answer is:
are in H.P.
are in H.P.
$\left|\begin{array}{lll}1 & 2 \mathrm{a} & \mathrm{a} \\ 1 & 3 \mathrm{~b} & \mathrm{~b} \\ 1 & 4 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{C}_2 \rightarrow \mathrm{C}_2-2 \mathrm{C}_3$
$\left|\begin{array}{ccc}1 & 0 & \mathrm{a} \\ 1 & \mathrm{~b} & \mathrm{~b} \\ 1 & 2 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$
$\left|\begin{array}{ccc}1 & 0 & a \\ 0 & b & b-a \\ 0 & 2 c-b & c-b\end{array}\right|=0$
$b(c-b)-(b-a)(2 c-b)=0$
On simplification,
$$
\frac{2}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}
$$
$\therefore \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in Harmonic Progression.
$\left|\begin{array}{ccc}1 & 0 & \mathrm{a} \\ 1 & \mathrm{~b} & \mathrm{~b} \\ 1 & 2 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$
$\left|\begin{array}{ccc}1 & 0 & a \\ 0 & b & b-a \\ 0 & 2 c-b & c-b\end{array}\right|=0$
$b(c-b)-(b-a)(2 c-b)=0$
On simplification,
$$
\frac{2}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}
$$
$\therefore \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in Harmonic Progression.
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