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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
Solution
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
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