RuslanMRF wrote:
Series 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 is possible?
I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers
I only
II only
I and III only
II and III only
I, II, and III
Please, explain the the solution.
New edition of this question with a solution:
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
Given that the sequence of numbers \(a_1\), \(a_2\), \(a_3\), ... have the following properties: \(i*a_i=j*a_j\) and \(a_1=positive \ integer\), so \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \ integer\).
We should determine whether the options given below can occur (notice that the question is which of the following
COULD be true, not MUS be true).
I. \(2a_{100}=a_{99}+a_{98}\) --> since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\) --> reduce by \(a_{100}\) --> \(2=\frac{100}{99}+\frac{100}{98}\) which is not true. Hence this option
could be true.
II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be non-integers --> \(a_1=1=2a_2=3a_3=...\) --> \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option
could be true.
III. The sequence does not contain negative numbers --> since given that \(a_1=positive \ integer=n*a_n\), then \(a_n=\frac{positive \ integer}{n}=positive \ number\), hence this option is
always true.
Answer: D.
Hope it's clear.
can you please explain me option A. i am totally confused with it