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Question

If ∣ ∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣ ∣=0 then show that a,b,c are in A. P.

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Solution

∣ ∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣ ∣=0
Using the row transformation, R1R2R1 and R2R3R2, we get
∣ ∣11ba11cbx+3x+4x+c∣ ∣
and then using R1R2R1 we get,
∣ ∣002bac11cbx+3x+4x+c∣ ∣.
Now expanding the determinant by R1 we get (2bac)×1=0.
So, 2b=a+c.
Hence, a,b,c are in A.P

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