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In the sequence above each term after

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In the sequence above each term after  [#permalink]

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New post Updated on: 30 Jun 2017, 02:39
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1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

Originally posted by HarveyKlaus on 10 Jul 2016, 01:35.
Last edited by abhimahna on 30 Jun 2017, 02:39, edited 1 time in total.
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Re: In the sequence above each term after  [#permalink]

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New post 10 Jul 2016, 05:23
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HarveyKlaus wrote:
Hi all,
I am not able to solve this question by the normal sequence formula. Can someone please help? Thanks

1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001


Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. \(2^1\), \(2^2\),\(2^3\).... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.
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Re: In the sequence above each term after  [#permalink]

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New post 10 Jul 2016, 05:39
It's just about calculating the tenth term which is 2^10=1024.
Then you should be able to define where this number fits.
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Re: In the sequence above each term after  [#permalink]

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New post 10 Jul 2016, 06:31
In the Sequence notice that the sequence is just the 1/(2^n) ...
so for 1st term=1/2^1=1/2
2nd term=1/(2^2)1/4, 3rd term=1/(2^3)=1/8 and so on...
Thus the 10th term will be 1/(2^10)=1/1024
Roughly, 1/1024 can be 1/1000=0.001 but since denominator is a bit more than 1000 therefore the actual value will be a bit less than 0.001.

thus the ans will lie btw. 0.0001 and 0.001.(D)
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Re: In the sequence above each term after  [#permalink]

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New post 03 Jun 2017, 12:23
msk0657 wrote:
Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. \(2^1\), \(2^2\),\(2^3\).... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.


What is your approach to solve for 2^10 and 1/1024?
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Re: In the sequence above each term after  [#permalink]

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New post 06 Jun 2017, 12:04
msk0657 wrote:
HarveyKlaus wrote:

1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001


Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. \(2^1\), \(2^2\),\(2^3\).... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.


Then 1/1024 = 0.024
msk0657 , how did you get this value?

On a calculator it comes out as .0009765625.
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Re: In the sequence above each term after  [#permalink]

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New post 10 Sep 2018, 04:10
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