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 nonzero solution then a, b, c

are in GP

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are in AP

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are in HP

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satisfy a+2b+3c= 0

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Solution

The correct option is B are in HP
System of equations has non-zero solution.
i.e $Δ = 0$
Applying $R_{3} \to R_{3} - R_{1}, R_{2} \to R_{2} - R_{1}$
$∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 a & a 1 & 3 b & b 1 & 4 c & c \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 a & a 0 & 3 b - 2 a & b - a 0 & 4 c - 2 a & c - a \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 2 a c = a b + b c$
Hence $a, b, c$ are in HP

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Similar questions

x + 2 a y + a z = 0

x + 3 b y + b z = 0

x + 4 c y + c z = 0

a, b, c

x + 2 a y + a z = 0

x + 3 b y + b z = 0

x + 4 c y + c z = 0

a, b, c

x + 2 a y + a z = 0, x + 3 b y + b z = 0,

x + 4 c y + c z = 0

x + 2 a y + a z = 0, x + 3 b y + b z = 0, x + 4 c y + c z = 0

x + 2 a y + a z = 0, x + 3 b y + b z = 0, x + 4 c z + c z = 0

a, b, c

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