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Question

If x,y,z are in A.P., then the value of the determinant ∣ ∣a+2a+3a+2xa+3a+4a+2ya+4a+5a+2z∣ ∣ is

A
1
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B
0
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C
2a
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D
a
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Solution

The correct option is B 0
Since, a,b,c are in A.P.
2b=a+c
∣ ∣a+2a+3a+2xa+3a+4a+2ya+4a+5a+2z∣ ∣
R1R1+R3
=∣ ∣ ∣2(a+3)2(a+4)2a+2(x+z)a+3a+4a+2ya+4a+5a+2z∣ ∣ ∣
=2∣ ∣ ∣(a+3)(a+4)a+2ya+3a+4a+2ya+4a+5a+2z∣ ∣ ∣
=0

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