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

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Difficulty:   45% (medium)

Question Stats: 70% (01:55) correct 30% (02:04) wrong based on 1344 sessions

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

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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: $$0.0001 < a_{10}< 0.001$$;

$$0.0001 < 4a_{12}< 0.001$$;

$$0.000025 < a_{12}< 0.00025$$.

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

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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

##### General Discussion
Intern  Joined: 18 Sep 2016
Posts: 43
Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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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

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

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Bunuel and VeritasPrepKarishma .. Could you please help here sharing the approach? Stuck in this Senior Manager  G
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Re: In a certain sequence, each term after the first term is one-half the  [#permalink]

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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, 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"
Senior Manager  G
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In a certain sequence, each term after the first term is one-half the  [#permalink]

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

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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

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

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

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

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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. Re: In a certain sequence, each term after the first term is one-half the   [#permalink] 18 Jan 2020, 05:20
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