Question

`|(x + 1, x + 2, x + "a"),(x + 2, x + 3, x + "b"),(x + 3, x + 4, x + "c")|` = 0, where a, b, c are in A.P.

Options

• True

• False

Answer 1

This statement is True.

Explanation:

Let Δ = `|(x + 1, x + 2, x + "a"),(x + 2, x + 3, x + "b"),(x + 3, x + 4, x + "c")|`

R₂ → 2R₂ – (R₁ + R₃)

= `|(x + 1, x + 2, x + "a"),(0, 0, 2"b" - ("a" + "c")),(x + 3, x + 4, x + "c")|`

a, b, c are in A.P.

∴ b – a = c – b

⇒ 2b = a + c

= `|(x + 1, x + 2, x + "a"),(0, 0, 0),(x + 3, x + 4, x + "c")|`

= 0