Let a, b, c be such that b(a+c)=0. If ∣aa+1a−1−+1b−1−1c+1∣+∣∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣∣=0then the value of n is

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Let a, b, c be such that {b}{({a}{}+{}{c})}{0}. If {|{a}{a}+{1

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Let a, b, c be such that $b (a + c) \neq = 0$ . If $∣ a a + 1 a - 1 - + 1 b - 1 - 1 c + 1 ∣ + ∣ ∣ a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 (- 1)^{n + 2} a (- 1)^{n + 1} b (- 1)^{n} c ∣ ∣ = 0$ then the value of n is

zero

any even integer

any odd integer

any integer

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Amelia Discussed

Let a, b, c be such that

b (a + c) \neq = 0

. If

∣ a a + 1 a - 1 - + 1 b - 1 - 1 c + 1 ∣ + ∣ ∣ a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 (- 1)^{n + 2} a (- 1)^{n + 1} b (- 1)^{n} c ∣ ∣ = 0

then the value of n is

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Question Text	Let a, b, c be such that $b (a + c) \neq = 0$ . If $∣ a a + 1 a - 1 - + 1 b - 1 - 1 c + 1 ∣ + ∣ ∣ a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 (- 1)^{n + 2} a (- 1)^{n + 1} b (- 1)^{n} c ∣ ∣ = 0$ then the value of n is
Answer Type	Video solution: 1
Upvotes	150