In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 : GMAT Data Sufficiency (DS)
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# In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1

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

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18 Jun 2008, 09:23
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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 $$t_5$$?

(1) $$t_3 = \frac{1}{4}$$

(2) $$t_1 - t_5 = \frac{15}{16}$$
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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18 Jun 2008, 11:11
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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:

15/16=t1-t5=t1-t1*q^4 = t1*(1-q^4) = t1*(1-1/16) = t1*15/16.

So, from here it follows that t1=1 and t5=t1*q^4 = 1*1/16 = 1/16.

Sufficient.

I hope that helped.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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18 Jun 2008, 12:43
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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

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

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19 Dec 2010, 15:31
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.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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19 Dec 2010, 15:42
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gettinit wrote:
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.

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.

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

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23 Dec 2010, 17:42
Thanks for the explanation Bunuel. I can see it now clearly.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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15 Apr 2014, 01:43
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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10 May 2014, 11:12
statement(1)

tn+1 = tn/2
t3 = 1/4
t4 =t3/2 = 1/8
t5 = t4/2 = 1/16. Sufficient.

statement (2)
t1 - t5 = 15/16
t1 = t5 + 15/16 = (16t5 + 15) /16
t2 = t1/2 = (16t5 + 15) / 32
t3 = t2/2 = (16t5 + 15) /64
t4 = t3/2 = (16t5 + 15) /128
t5 = t4/2 = (16t5 + 15) /256

256t5 = 16t5 + 15
240t5 = 15
t5 = 15/240 = 5/80 = 1/16
Sufficient.

Thus, D.
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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, 09:31
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
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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23 Nov 2014, 04:33
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deeuk wrote:
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, 06:56
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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, 21:25
pls change the format of the question its a bit confusing.

the entire expression seems equaled to tn/2
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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, 21:38
hsbinfy wrote:
pls change the format of the question its a bit confusing.

the entire expression seems equaled to tn/2

______________
Edited. Thank you.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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17 Sep 2016, 09:38
I am having trouble understanding how you went from T3 = T2/2 = T1/4

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

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17 Sep 2016, 09:57
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g3lo18 wrote:
I am having trouble understanding how you went from T3 = T2/2 = T1/4

Notice that we are given T(n+1) = T(n)/2

So, substitute n = 2, you will get T(3) = T(2)/2.

Now substitute n =1 , you will get T(2) = T(1)/2

So, we can say T(3) = T(2)/2 = T(1)/4
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]

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17 Sep 2016, 10:01
OMG I see it now. Thank you. Wow these are so annoying to deal with. Guess I need to get accustomed to it.
Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1   [#permalink] 17 Sep 2016, 10:01
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