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Difficulty: 555-605 Level,   Fractions and Ratios,   Inequalities,   Sequences,                              
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
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]
Bunuel and VeritasPrepKarishma .. Could you please help here sharing the approach? Stuck in this :(
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
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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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

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

Since the tenth term has 3 to 4 values after the decimal point, the eleventh term will have 4 to 5 values and the 12th will have 5 to 6. Since 25 takes two places, the zeroes in front will be either three or four as in answer C?
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
kokus wrote:
Is the following approach correct:

Since the tenth term has 3 to 4 values after the decimal point, the eleventh term will have 4 to 5 values and the 12th will have 5 to 6. Since 25 takes two places, the zeroes in front will be either three or four as in answer C?

risky to assume. I will do it this way -

10th term: 0.0001 TO 0.001
11th term: 0.00005 TO 0.0005 (divide the above by 2)
12th term: 0.000025 TO 0.00025 (divide the above by 2)
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
AnirudhaS wrote:
kokus wrote:
Is the following approach correct:

Since the tenth term has 3 to 4 values after the decimal point, the eleventh term will have 4 to 5 values and the 12th will have 5 to 6. Since 25 takes two places, the zeroes in front will be either three or four as in answer C?

risky to assume. I will do it this way -

10th term: 0.0001 TO 0.001
11th term: 0.00005 TO 0.0005 (divide the above by 2)
12th term: 0.000025 TO 0.00025 (divide the above by 2)


Thanks, but why is it risky? You just need to be aware what the product will be and then just add the number of zeroes?
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In a certain sequence, each term after the first term is one-half the [#permalink]
Given: In a certain sequence, each term after the first term is one-half the previous term.
Asked: If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between

Let the first term of the sequence be x
\(T_1 = x\)
\(T_2 = \frac{x}{2}\)
\(T_n = \frac{x}{2^{n-1}}\)

\(T_{10} = \frac{x}{2^9}\);
\(.0001 < \frac{x}{2^9} < .001\)

\(T_{12} = \frac{x}{2^{11}}\)
\(\frac{.0001}{4} < T_{12} < \frac{.001}{4}\)
\(.000025 < T_{12} < .00025\)

IMO C
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In a certain sequence, each term after the first term is one-half the [#permalink]
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10th term = between 0.0001 and 0.001

Convert to fraction: \(\frac{1}{10000}\) and \(\frac{1}{1000}\)

11th term = between \(\frac{1}{20000}\) and \(\frac{1}{2000}\)

12th term = between \(\frac{1}{40000}\) and \(\frac{1}{4000}\)

12th term = between \(\frac{1}{4} * \frac{1}{10000}\) and \(\frac{1}{4} * \frac{1}{1000}\)

12th term = between \(0.000025\) and \(0.00025\)
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In a certain sequence, each term after the first term is one-half the [#permalink]
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Instead of 0.0001 and 0.001, get rid of the decimals by multiplying the two numbers by 10^4 which gives us 1 and 10 (we'll put the decimals back later by multiplying the results by 10^-4)

1*1/2= 0,5 and 10*1/2=5
0,5*1/2=0,25 and 5*1/2=2,5

Result is : 0,25*10^-4= 0,000025 and 2,5*10^-4 = 0,00025

Answer C)
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
Can we just say that the exponent is the number of zeros between the decimal and the first non-zero digit?
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In a certain sequence, each term after the first term is one-half the [#permalink]
Hi CEdward,

Not really.

Here are a few examples :
125/1000= 125*10^-3 = 0,125 Here there are no zeros

12/10000 = 12*10^-4 = 0,0012 Here there are 2 zeros between the decimal and the first non zero integer, and the exponent is -4

CEdward wrote:
Can we just say that the exponent is the number of zeros between the decimal and the first non-zero digit?
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Re: In a certain sequence, each term after the first term is one-half the [#permalink]
Sabby wrote:
Hi CEdward,

Not really.

Here are a few examples :
125/1000= 125*10^-3 = 0,125 Here there are no zeros

12/10000 = 12*10^-4 = 0,0012 Here there are 2 zeros between the decimal and the first non zero integer, and the exponent is -4

CEdward wrote:
Can we just say that the exponent is the number of zeros between the decimal and the first non-zero digit?


Thanks! What's the fastest way to do such divisions without long division?

Posted from my mobile device
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