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

a+2a+3a+2xa+3a+4a+2ya+4a+5a+2z R1→R1+R3 2a+62a+82a+2x+2za+3a+4a+2ya+4a+5a+2z ≈2a+3a+4a+2ya+3a+4a+2ya+4a+5a+2z=0

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

a+2a+3a+2xa+3a+4a+2ya+4a+5a+2z R1→R1+R3 2a+62a+82a+2x+2za+3a+4a+2ya+4a+5a+2z ≈2a+3a+4a+2ya+3a+4a+2ya+4a+5a+2z=0

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If x, y, z are in A.P. then the value of $|\begin{matrix} a + 2 & a + 3 & a + 2 x \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{matrix}| =$

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answer is B.

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Detailed Solution

$|\begin{matrix} a + 2 & a + 3 & a + 2 x \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{matrix}| R_{1} \to R_{1} + R_{3} |\begin{matrix} 2 a + 6 & 2 a + 8 & 2 a + 2 x + 2 z \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{matrix}| \approx 2 |\begin{matrix} a + 3 & a + 4 & a + 2 y \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{matrix}| = 0$