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In the sequence above each term after
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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 onehalf 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
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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.
Edited the question



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Re: In the sequence above each term after
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10 Jul 2016, 05:23
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 onehalf 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
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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
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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
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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
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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 onehalf 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 = 0.024msk0657 , 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
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16 Jan 2019, 15:21
HarveyKlaus wrote: 1/2, 1/4, 1/8, 1/16, 1/32, ....
In the sequence above each term after after the first onehalf 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 When comparing powers of some value to either powers of 10 or some decimal with a trailing 1 (which is also a power of 10), one approach is to rewrite our term, to get it as close to a power of 10 as possible. For (1/2)^10, a base of 1/8 is gonna be as close to 1/10 as we can get. Then (1/2)^10 = 1/(2^3 *2^3 *2^3 *2) = (1/8)^3*1/2 is approximately (1/10)^3 *1/2 = .001*1/2 = approximately .0005, which falls between .0001 and .001. The answer is D



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In the sequence above each term after
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20 Jan 2019, 10:50
msk0657 wrote: 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 onehalf 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. Hello msk0657 ! How did you get this result? Then 1/1024 = 0.024 Kind regards!




In the sequence above each term after
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