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in Binomial theorem by (46.7k points)

Let a1 = 0 and a1, a2, a3, …, an be real numbers such that |ai| = |ai − 1 + 1| for all i = 0, 1, 2, …, n. If the AM of the numbers a1, a2, a3, …, an has the value x, then

(A)  x < 1

(B)  x <  1/2

(C)  x ≥ - 1/2

(D)  x ≥ 1

1 Answer

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by (52.7k points)
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Best answer

Correct option  (C)  x ≥ - 1/2

|ai| = |ai - 1 + 1| for i = 2,............,n

Squaring we have a2i = a2i - 1 + 2ai - 1 + 1 ⇒ a2i - a2i - 1 = 2ai - 1 + 1

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