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If x1, x2, x3… xn is a sequence such that xn=n/2 for all values x≥1, i

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If x1, x2, x3… xn is a sequence such that xn=n/2 for all values x≥1, i [#permalink]

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If \(x_1\), \(x_2\), \(x_3\), … \(x_n\) is a sequence such that \(x_n=\frac{n}{2}\) for all values \(x \geq 1\), is \(x_b\) less than \(x_h\) ?


(1) \(\sqrt{b} \geq \sqrt{h}\)

(2) b and h are distinct prime numbers.
[Reveal] Spoiler: OA

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Re: If x1, x2, x3… xn is a sequence such that xn=n/2 for all values x≥1, i [#permalink]

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New post 28 Sep 2017, 05:21
Bunuel wrote:
If \(x_1\), \(x_2\), \(x_3\), … \(x_n\) is a sequence such that \(x_n=\frac{n}{2}\) for all values \(x \geq 1\), is \(x_b\) less than \(x_h\) ?


(1) \(\sqrt{b} \geq \sqrt{h}\)

(2) b and h are distinct prime numbers.


eventually the question is whether b>h
a)considering b=1 h=1 1<1 No
b=2 h=1 2>1 yes...Insuff

b)b and can be any prime no,making either of the two possibilities of b>h or h>b..insuff

c)makes sense

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If x1, x2, x3… xn is a sequence such that xn=n/2 for all values x≥1, i [#permalink]

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New post 28 Sep 2017, 07:25
Bunuel wrote:
If \(x_1\), \(x_2\), \(x_3\), … \(x_n\) is a sequence such that \(x_n=\frac{n}{2}\) for all values \(x \geq 1\), is \(x_b\) less than \(x_h\) ?


(1) \(\sqrt{b} \geq \sqrt{h}\)

(2) b and h are distinct prime numbers.


To find \(x_b\)\(<\)\(x_h\)

or \(\frac{b}{2}<\frac{h}{2}=> b<h\)

Statement 1: as \(b\) & \(h\) are positive, we can directly square the inequality to get \(b≥h\)

Now if \(b>h\), then the answer to our question stem is NO

and if \(b=h\), then also answer to our question stem is NO. Hence Sufficient

Statement 2: \(b\) and \(h\) can be any prime nos and hence \(b>h\) or \(b<h\). Hence Insufficient

Option A

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If x1, x2, x3… xn is a sequence such that xn=n/2 for all values x≥1, i   [#permalink] 28 Sep 2017, 07:25
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