The sequence a_1 , a_2 , a_3 , ..., a_n , ... is such that i*a_i=j*a_j for any pair of positive integers (i, j) . If a_1 is a positive integer, which of the following could be true? I. 2*a_{100}=a_{99}+a_{98} II. a_1 is the only integer in the sequence III. The sequence does not contain negative numbers A. I only B. II only C. I and III only D. II and III only E. I, II, and III

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VidhiJ

Joined: 03 Dec 2023

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27 Dec 2025, 03:04

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2*a100 can we written as a100 + a100. According to the question 100*a100 = 99*a99 and 100*a100 = 98*a98. Since the question mentions it's true for every pair of interger (i,j), we don't need to think of a sequence. Anyway, 1, 1/2, 1/3, 1/4 satisfies the condition

Bunuel

Your scenario will occur when \(a_1=1\). In this case, the sequence will indeed be {1, 1/2, 1/3, ..., 1/98, 1/99, 1/100}. Note that for any term in the sequence, the product of that term and its index (its position number in the sequence) is always 1: \(1*a_1=2*a_2=3*a_3=...=100*a_{100}=1\). However, \(2*a_{100}=a_{99}+a_{98}\) does not hold as \(2*a_{100}=\frac{1}{50}\), whereas \(a_{99}+a_{98}=\frac{1}{98}+\frac{1}{99}=\frac{197}{9702}\neq 2\).

Furthermore, observe that \(i*a_1\) is not always 1. For example, when \(a_1=2\), the sequence becomes {2, 2/2, 2/3, 2/4, 2/5, 2/6, ...} and in this scenario, \(i*a_1=2\), not 1.

Hope it helps.