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In a certain sequence, each term after the first term is one-half the

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New post 27 May 2016, 03:46
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In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

(A) 0.0025 and 0.025
(B) 0.00025 and 0.0025
(C) 0.000025 and 0.00025
(D) 0.0000025 and 0.000025
(E) 0.00000025 and 0.0000025
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 10 Dec 2017, 03:19
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sadikabid27 wrote:
Bunuel and VeritasPrepKarishma .. Could you please help here sharing the approach? Stuck in this :(


In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

(A) 0.0025 and 0.025
(B) 0.00025 and 0.0025
(C) 0.000025 and 0.00025
(D) 0.0000025 and 0.000025
(E) 0.00000025 and 0.0000025

Each term after the first term is one-half the previous term: \(a_{n+1}=\frac{a_n}{2}\). So:

\(a_{11}=\frac{a_{10}}{2}\)

\(a_{12}=\frac{a_{11}}{2}=\frac{a_{10}}{4}\) --> \(4a_{12}=a_{10}\).


Given: \(0.0001 < a_{10}< 0.001\);

\(0.0001 < 4a_{12}< 0.001\);

\(0.000025 < a_{12}< 0.00025\).

Answer: C.
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New post 27 May 2016, 04:11
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12th Term will be obtained by multiplying the 10th term by \(\frac{1}{2}\) * \(\frac{1}{2}\)

Let 10th term is a

0.0001 < \(a_{10}\) < 0.001

12th term will be

\(\frac{0.0001}{4}\) < \(a_{12}\) < \(\frac{0.001}{4}\)

0.000025 < \(a_{12}\) < 0.00025

Option C is the answer
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 28 Sep 2016, 02:13
the low limit of the number is equal to 1/10 of the highter limit of the number.

if you divide the low limit you'll find 0.000025

as a consequence the high limit of the number will be the low one *10 = 0.00025

answer C
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In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 10 Dec 2017, 01:49
Bunuel and VeritasPrepKarishma .. Could you please help here sharing the approach? Stuck in this :(
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New post 16 Jun 2018, 09:41
Multiply original number 0.0001 and 0.001 with 1,00,000 both, you get:

10 and 100, divide both by 2, 5 and 50 further divide by 2, 2.5 and 25, now divide by 1,00,000 or push both numbers by five zeros, you get "C"
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New post 16 Jun 2018, 09:43
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Multiply original number 0.0001 and 0.001 with 1,00,000 both, you get:

10 and 100, divide both by 2, you get 5 and 50,
further divide 5 and 50, by 2, you get 2.5 and 25,

now divide by 1,00,000 or push both numbers by five zeros, you get "C". Tadaaaaaaaaaaaa!
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New post 25 Aug 2019, 08:12
nalinnair wrote:
In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

(A) 0.0025 and 0.025
(B) 0.00025 and 0.0025
(C) 0.000025 and 0.00025
(D) 0.0000025 and 0.000025
(E) 0.00000025 and 0.0000025


Convert 0.0001 and 0.001 into scientific notation
0.0001 = 1* 10^-4
0.001 = 1* 10^-3

Because each term is 1/2 of the previous term, \(Term 12 = \frac{1}{2}*\frac{1}{2}*Term 10\)

\(Term 12 = \frac{1}{4}*Term 10\)

(.25* 10^-4) = \(\frac{1}{4}\)*(1* 10^-4)

(.25* 10^-3) = \(\frac{1}{4}\)*(1* 10^-3)

Then move the decimal place 4 and 3 spots to the left, which gives you 0.000025 and 0.00025

Answer is C
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 26 Aug 2019, 15:15
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nalinnair wrote:
In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

(A) 0.0025 and 0.025
(B) 0.00025 and 0.0025
(C) 0.000025 and 0.00025
(D) 0.0000025 and 0.000025
(E) 0.00000025 and 0.0000025


\(\frac{1}{10000} < n_{10} < \frac{1}{1000}\)

\((\frac{1}{2^{2}}) * \frac{1}{10000} < n_{12} < (\frac{1}{2^{2}})* \frac{1}{1000}\)

\(\frac{1}{4} * 10^{-4} < n_{12} < \frac{1}{4} * 10^{-3}\)

\(.25 * 10^{-4} < n_{12} < .25 * 10^{-3}\)

\(.000025 < n_{12} < .00025\)
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 17 Jan 2020, 11:37
In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

(A) 0.0025 and 0.025
(B) 0.00025 and 0.0025
(C) 0.000025 and 0.00025
(D) 0.0000025 and 0.000025
(E) 0.00000025 and 0.0000025

12th term would be 10th term divided by 4 as each consecutive term is one half of the previous term.

a10/4 = 0.0001/4

1/10000*4

1/10000 * 1/4

1/10000 * 0.25

For four 0's move the decimal four places to the left.

.000025

Only option (C) matches
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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New post 18 Jan 2020, 05:20
Alternate solution, 12th term means: you need to divide the number twice by 2 i.e by 4 (1/2*1/2).

when you divide 0.001/4 you get 0.00025; easiest way.
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