yeahwill 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
a. I only
b. II only
c. I and III only
d. II and III only
e. I, II, and III
II is possible. If A(1)=1 then A(k) = \frac{1}{k} . There are no more integers in the sequence except A(1) .
III is possible as well. Same example applies.
I is not possible. Because i*A(i) = (i + 1)A(i + 1)A(i + 1) = \frac{i}{i + 1} A(i) which is less then A(i). This means that this sequence is a decreasing sequence in which every subsequent element is smaller than its predecessor. Thus, A(100) + A(100) \lt A(99) + A(98)
The correct answer is D.
Can someone please explain this?
I have no clue what data is provided and what answer is expected in this question
This question was posted in PS forum. Below is my post from there:
A set 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 (note that the question is which
can be true, not must be true).
I.
2a_{100}=a_{99}+a_{98} --> as
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
cannot be true.
II.
a_1 is the only integer in the series. 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
can be true.
III. The series does not contain negative numbers --> as 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 (II and III only).
Hope it's clear.
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