Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 May 2015, 12:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 14 May 2008
Posts: 71
Followers: 1

Kudos [?]: 36 [0], given: 0

In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]  18 Jun 2008, 09:23
3
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

64% (02:19) correct 36% (01:34) wrong based on 161 sessions
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
[Reveal] Spoiler: OA
Director
Joined: 12 Apr 2008
Posts: 500
Location: Eastern Europe
Schools: Oxford
Followers: 12

Kudos [?]: 177 [1] , given: 4

Re: DS T [#permalink]  18 Jun 2008, 11:11
1
KUDOS
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.
Manager
Joined: 11 Apr 2008
Posts: 129
Location: Chicago
Followers: 1

Kudos [?]: 39 [5] , given: 0

Re: DS T [#permalink]  18 Jun 2008, 12:43
5
KUDOS
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
Attachments

sequence.gif [ 6.83 KiB | Viewed 6264 times ]

_________________

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

Manager
Joined: 13 Jul 2010
Posts: 169
Followers: 1

Kudos [?]: 30 [0], given: 7

Re: DS T [#permalink]  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.
Math Expert
Joined: 02 Sep 2009
Posts: 27542
Followers: 4330

Kudos [?]: 42660 [0], given: 6052

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

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.
_________________
Manager
Joined: 13 Jul 2010
Posts: 169
Followers: 1

Kudos [?]: 30 [0], given: 7

Re: DS T [#permalink]  23 Dec 2010, 17:42
Thanks for the explanation Bunuel. I can see it now clearly.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4979
Followers: 300

Kudos [?]: 55 [1] , given: 0

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

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 08 Jan 2014
Posts: 20
Location: United States
Concentration: General Management, Entrepreneurship
GMAT Date: 06-30-2014
GPA: 3.99
WE: Analyst (Consulting)
Followers: 1

Kudos [?]: 14 [0], given: 4

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]  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.
Intern
Joined: 23 Aug 2014
Posts: 43
GMAT Date: 11-29-2014
Followers: 0

Kudos [?]: 7 [0], given: 28

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]  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
Math Expert
Joined: 02 Sep 2009
Posts: 27542
Followers: 4330

Kudos [?]: 42660 [1] , given: 6052

Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]  23 Nov 2014, 04:33
1
KUDOS
Expert's post
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.
_________________
Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1   [#permalink] 23 Nov 2014, 04:33
Similar topics Replies Last post
Similar
Topics:
1 In the sequence of non-zero numbers t1, t2,t3,......tn,..... 4 12 Jul 2008, 21:34
In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 5 10 Jul 2008, 22:49
In the arithmetic sequence t(1), t(2), t(3)....t(n). 5 25 Mar 2007, 07:42
In the arithmetic sequence, t1, t2, t3, ........, tn, 3 25 Sep 2006, 21:28
1 If the terms of a sequence are t1,t2,t3...tn what is the the 4 07 May 2006, 17:08
Display posts from previous: Sort by