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In the sequence 1, 2, 4, 8, 16, 32, …, each term after the
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03 Nov 2010, 02:51
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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 4Q15 concept question.
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30 Jan 2014, 21:33
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)^n1 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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In the sequence 1, 2, 4, 8, 16, 32, …, each term after the
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03 Nov 2010, 03:14
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 \(\frac{1}{2}\): \(t_1\), \(\frac{t_1}{2}\), \(\frac{t_1}{4}\), \(\frac{t_1}{8}\), ... > \(t_n=\frac{t_1}{2^{n1}}\). 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) \(t_1t_5=2^4*t_5t_5=\frac{15}{16}\) > 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. Hope it helps.
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Re: Sequence of NonZero
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03 Nov 2010, 03:19
Thanks Bunuel for the detailed explanation.
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GWD set 4Q15 concept question.
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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)^n1 And I don't understand how this works. Can some explain how this concept work??



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Re: GWD set 4Q15 concept question.
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30 Jan 2014, 20:27
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)^n1 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)^n1. In this theory they just generalize it. Given  tn+1 = tn/2 Putting n = n 1 tn= t(n1) *1/ 2  consider this first equation. while t(n1) = t(n2)*1/2  consider this 2nd equation. if you put t(n1) in first equation tn = t(n2) * 1/ 2^2 so when we need t1 on the right hand side. tn=t1*(1/2)^n1 Easy way to understand is  t2 = t1/2 t3 = t2 / 2 t3 = t1* 1/2^2 So on generalization tn = t1* 1/ 2^(n1) Hope I make it clear.
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Re: GWD set 4Q15 concept question.
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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)^n1 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)^n1. In this theory they just generalize it. Given  tn+1 = tn/2 Putting n = n 1 tn= t(n1) *1/ 2  consider this first equation. while t(n1) = t(n2)*1/2  consider this 2nd equation. if you put t(n1) in first equation tn = t(n2) * 1/ 2^2 so when we need t1 on the right hand side. tn=t1*(1/2)^n1 Easy way to understand is  t2 = t1/2 t3 = t2 / 2 t3 = t1* 1/2^2 So on generalization tn = t1* 1/ 2^(n1) Hope I make it clear. Thanks! I considered your explanation and got the idea.



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Re: GWD set 4Q15 concept question.
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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)^n1 And I don't understand how this works. Can some explain how this concept work?? Merging similar topics. Please refer to the solutions above. Also, please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Pay attention to rules 1 and 3. Thank you.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the
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