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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] New post 18 Jun 2008, 09:23
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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: DS T [#permalink] New post 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: DS T [#permalink] New post 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


see attached
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Re: DS T [#permalink] New post 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: DS T [#permalink] New post 19 Dec 2010, 15:42
Expert's post
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.

Answer: D.

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

- any particular two terms,
- any particular term and common difference (common ratio),
- any particular term and the formula for n_th term,
- 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: DS T [#permalink] New post 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] New post 15 Apr 2014, 01:43
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink] New post 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.
Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1   [#permalink] 10 May 2014, 11:12
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