If a, b, c are in AP, then <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>x</mi><mo>+</mo><mn>1</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>a</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mn>2</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>b</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mn>3</mn></mtd><mtd><mi>x</mi><mo>+</mo><mn>4</mn></mtd><mtd><mi>x</mi><mo>+</mo><mi>c</mi></mtd></mtr></mtable></mfenced></math> equals -
Step-by-Step Solution
Key Concept: Since a, b, c are in AP, 2b = a + c. Applying row operations R1 -> R1 + R3 - 2R2 will make the third column zero.
Given a, b, c are in AP, we have 2b = a + c. Let the determinant be \Delta. Applying R1 -> R1 + R3, we get the first row as (2x+4, 2x+6, 2x+a+c). Since a+c = 2b, the third element is 2x+2b. Now, R1 -> R1 - 2R2 gives (2x+4 - 2(x+2), 2x+6 - 2(x+3), 2x+2b - 2(x+b)) = (0, 0, 0). Since one row is zero, the determinant is 0.
Correct Answer: 3