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In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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18 Jun 2008, 10:23

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In the sequence of nonzero numbers \(t_1\), \(t_2\), \(t_3\), …, \(t_n\), …, \(t_{n+1}=\frac{t_n}{2}\) for all positive integers n. What is the value of \(t_5\)?

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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18 Jun 2008, 12:11

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Hello, quantum, this is my attempt to explain why it's D:

Quote:

In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers n. What is the value of t5? (1) t3 = 1/4 (2) t1 - t5 = 15/16

Here we have geometric progression, i.e. series where t2=t1*q, t3=t2*q, …, tn+1=tn*q. In our case, q=0.5. Also note that tn+1=t1*q^n

So, basically, to answer this question, it is sufficient to know the value of any of the tn.

1) Explicitly gives us the value for t3, so it’s sufficient.

2) So, let’s see if we can obtain the value of t1 from this statement, using the formula tn+1=t1*q^n:

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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18 Jun 2008, 13:43

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quantum wrote:

In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers n. What is the value of t5? (1) t3 = 1/4 (2) t1 - t5 = 15/16

see attached

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sequence.gif [ 6.83 KiB | Viewed 15108 times ]

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Factorials were someone's attempt to make math look exciting!!!

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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19 Dec 2010, 16:31

Good explanations but I got confused at how you equated tn+1=tn/2? I thought the it was the entire expression that equaled to tn/2? sorry but I am a bit confused. Thanks.

Good explanations but I got confused at how you equated tn+1=tn/2? I thought the it was the entire expression that equaled to tn/2? sorry but I am a bit confused. Thanks.

In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers n. What is the value of t5?

Given: \(t_{n+1}=\frac{t_n}{2}\). So \(t_2=\frac{t_1}{2}\), \(t_3=\frac{t_2}{2}=\frac{t_1}{4}\), \(t_4=\frac{t_3}{2}=\frac{t_1}{8}\), ...

Basically we have geometric progression with common ratio \(\frac{1}{2}\): \(t_1\), \(\frac{t_1}{2}\), \(\frac{t_1}{4}\), \(\frac{t_1}{8}\), ... --> \(t_n=\frac{t_1}{2^{n-1}}\).

Question: \(t_5=\frac{t_1}{2^4}=?\)

(1) \(t_3=\frac{1}{4}\) --> we can get \(t_1\) --> we can get \(t_5\). Sufficient. (2) \(t_1-t_5=2^4*t_5-t_5=\frac{15}{16}\) --> we can get \(t_5\). Sufficient.

Answer: D.

Generally for arithmetic (or geometric) progression if you know:

- any particular two terms, - any particular term and common difference (common ratio), - the sum of the sequence and either any term or common difference (common ratio),

then you will be able to calculate any missing value of given sequence.

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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15 Apr 2014, 02:43

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Hello from the GMAT Club BumpBot!

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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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22 Nov 2014, 10:31

Bunuel. I have a question about the 3rd generalized case. What does 'the formula for nth term' mean? Is it a+(n-1)d? I am guessing not since (a)we already know that as a formula and (b)Along with any particular term, the formula would still leave 2 variables- a and d. So are we talking about another equation for a term? Thanks again

Bunuel. I have a question about the 3rd generalized case. What does 'the formula for nth term' mean? Is it a+(n-1)d? I am guessing not since (a)we already know that as a formula and (b)Along with any particular term, the formula would still leave 2 variables- a and d. So are we talking about another equation for a term? Thanks again

You are right. I phrased that ambiguously. Will edit.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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11 Jul 2016, 07:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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