If the system of linear equation x+2ay+az=0,x+3by+bz=0,x+4cy+cz=0 has a non zero solution, then a, b, c
A
Are in A.P.
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B
Are in G. P.
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C
Are in H. P.
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D
Satisfy a +2b + 3c = 0
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Solution
The correct option is C Are in H. P. ∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c ∴ a, b, c are in Harmonic progression.