Since, a,b,c&d are in A.P., therefore
b−a=c−b=d−c=λ
We have,
∣x+a−cx−1x−b+dx+bx+cx+dx+ax+bx+c∣=2
C2→C2−C3
⇒∣x−2λx−1x+2λλλλx+ax+bx+c∣=2
R2→R2−R1,R3→R3−R1
∣x−2λ2λ−14λλ00x+ab−ac−a∣=2
⇒λ∣x−2λ2λ−14λ100x+aλ2λ∣=2
⇒−λ(4λ2−4λ2−2λ)=2
⇒λ2=1