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# In the sequence 1, 2, 4, 8, 16, 32, …, each term after the

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Manager
Joined: 16 Apr 2006
Posts: 225
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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03 Nov 2010, 02:51
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65% (01:15) correct 35% (01:32) wrong based on 295 sessions

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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: GWD set 4-Q15 concept question.  [#permalink]

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30 Jan 2014, 21:33
4
1
sehosayho 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

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??

This is the concept of a geometric progression (GP). In a GP, every term is related to the previous term by a fixed ratio r.
So $$t_2 = t_1*r$$
$$t_3 = t_2*r = t_1*r*r$$ (substituting from above)
$$t_4 = t_3*r = t_1*r*r*r$$
I hope you understand this.

Here, you are given that $$t_{n+1} = t_n *(\frac{1}{2})$$
So r = 1/2
$$t_2 = t_1 * (1/2)$$
$$t_3 = t_1*(1/2)*(1/2)$$
$$t_4 = t_1 *(1/2)^3$$
$$t_5 = t_1 * (1/2)^4$$

Statement 1 gives you $$t_1$$ so you get $$t_5$$ easily.
Statement 2 gives you $$t_1 - t_5$$. You can substitute $$t_5$$ from above and get $$t_1$$ and then $$t_5$$.

Check out this post on GPs: http://www.veritasprep.com/blog/2012/04 ... gressions/
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##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 49364
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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03 Nov 2010, 03:14
2
2
dkverma 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

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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03 Nov 2010, 03:19
Thanks Bunuel for the detailed explanation.
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Intern
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GWD set 4-Q15 concept question.  [#permalink]

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30 Jan 2014, 19:38
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

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??
Manager
Joined: 19 Apr 2013
Posts: 74
Concentration: Entrepreneurship, Finance
GMAT Date: 06-05-2015
GPA: 3.88
WE: Programming (Computer Software)
Re: GWD set 4-Q15 concept question.  [#permalink]

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30 Jan 2014, 20:27
1
sehosayho 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

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??

Hi, it is very simple.

The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1. In this theory they just generalize it.

Given -- tn+1 = tn/2

Putting n = n- 1

tn= t(n-1) *1/ 2 ----- consider this first equation.
while

t(n-1) = t(n-2)*1/2 ----- consider this 2nd equation.

if you put t(n-1) in first equation

tn = t(n-2) * 1/ 2^2

so when we need t1 on the right hand side.

tn=t1*(1/2)^n-1

Easy way to understand is ---

t2 = t1/2

t3 = t2 / 2

t3 = t1* 1/2^2

So on generalization

tn = t1* 1/ 2^(n-1)

Hope I make it clear.
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AB

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Posts: 11
Re: GWD set 4-Q15 concept question.  [#permalink]

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30 Jan 2014, 20:58
bhatiamanu05 wrote:
sehosayho 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

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??

Hi, it is very simple.

The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1. In this theory they just generalize it.

Given -- tn+1 = tn/2

Putting n = n- 1

tn= t(n-1) *1/ 2 ----- consider this first equation.
while

t(n-1) = t(n-2)*1/2 ----- consider this 2nd equation.

if you put t(n-1) in first equation

tn = t(n-2) * 1/ 2^2

so when we need t1 on the right hand side.

tn=t1*(1/2)^n-1

Easy way to understand is ---

t2 = t1/2

t3 = t2 / 2

t3 = t1* 1/2^2

So on generalization

tn = t1* 1/ 2^(n-1)

Hope I make it clear.

Thanks! I considered your explanation and got the idea.
Math Expert
Joined: 02 Sep 2009
Posts: 49364
Re: GWD set 4-Q15 concept question.  [#permalink]

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30 Jan 2014, 22:36
sehosayho 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

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??

Merging similar topics. Please refer to the solutions above.

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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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21 Feb 2017, 14:04
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the &nbs [#permalink] 21 Feb 2017, 14:04
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