GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 23 Jun 2018, 10:49

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In a geometric sequence each term is found by multiplying the previous

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46297
In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Jan 2015, 06:04
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

73% (00:45) correct 27% (00:42) wrong based on 175 sessions

HideShow timer Statistics

In a geometric sequence each term is found by multiplying the previous term by a constant. If the first and second terms in a geometric sequence are 2x and 4x, what is the 600th term of the sequence?


A. \(2^{599}*x\)

B. \(2^{600}*x\)

C. \(2^{(600x)}\)

D. \(4^{600}*x\)

E. \(4^{599}*x\)


Kudos for a correct solution.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Jan 2015, 07:08
hi the ans is B/C... as both B and C are the same
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46297
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Jan 2015, 07:19
Expert Post
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Jan 2015, 07:25
in this case ans is B...
constant term is 2..so 600th term is 2x*2^(600-1) which is same as x*2^600
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

1 KUDOS received
Manager
Manager
User avatar
Joined: 22 Oct 2014
Posts: 88
Concentration: General Management, Sustainability
GMAT 1: 770 Q50 V45
GPA: 3.8
WE: General Management (Consulting)
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Jan 2015, 07:36
1
Each element is found by multiplying the previous term with a constant, so we can start by dividing the second element by the first element:

\(\frac{4x}{2x}=2\)

Therefore, the constant that each previous term is multiplied with must be 2. Therefore:

\(a_3=8x\), m]a_4=16x[/m], etc.

We see that \(a_n=2^nx\)

So the 600th element is \(2^{600}x\)

The correct answer must be B.
_________________

\(\sqrt{-1}\) \(2^3\) \(\Sigma\) \(\pi\) ... and it was delicious!


Please consider giving +1 Kudos if deserved!

3 KUDOS received
SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 03 Feb 2015, 00:11
3
Answer = B. 2^600*x

\(1st term = 2x = 2^1 * x\)

2nd term \(= 4x = 2^2 * x\)

600th term \(= 2^{600} * x\)
_________________

Kindly press "+1 Kudos" to appreciate :)

Intern
Intern
avatar
Joined: 01 Feb 2015
Posts: 4
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 26 Mar 2015, 23:27
Isn't the formula for an explicit geometric sequence .... An = A1(r^(n-1))..

in my solution ... my answer was An = X*2^(599)
Because i subtracted the N = 600th term by 1. Please advise!

Thanks.
Expert Post
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 27 Mar 2015, 00:19
sisorayi01 wrote:
Isn't the formula for an explicit geometric sequence .... An = A1(r^(n-1))..

in my solution ... my answer was An = X*2^(599)
Because i subtracted the N = 600th term by 1. Please advise!

Thanks.


hi sisorayi01
you are absolutely correct about the formulae..
but A1 is 2x...
so 599 will again come to 600..
hope it helped
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

Intern
Intern
avatar
Joined: 01 Feb 2015
Posts: 4
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 27 Mar 2015, 08:26
Im sorry i still don't fully understand. How you get 600 for n-1 in the equation. Thank you.

Posted from my mobile device
Expert Post
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 5938
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 27 Mar 2015, 09:16
sisorayi01 wrote:
Im sorry i still don't fully understand. How you get 600 for n-1 in the equation. Thank you.

Posted from my mobile device



hi...
as the formula is An = A1(r^(n-1))..
A1=2x , r=2 and n=600...
An=2x*2^(600-1)= 2x*2^(599)=x*2^(600)...
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46297
Re: In a geometric sequence each term is found by multiplying the previous [#permalink]

Show Tags

New post 30 Dec 2017, 02:34
Bunuel wrote:
In a geometric sequence each term is found by multiplying the previous term by a constant. If the first and second terms in a geometric sequence are 2x and 4x, what is the 600th term of the sequence?


A. \(2^{599}*x\)

B. \(2^{600}*x\)

C. \(2^{(600x)}\)

D. \(4^{600}*x\)

E. \(4^{599}*x\)


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

The correct response is (B).

This abstract question asks for a very large term, so it’s a good opportunity to use the formula for a geometric sequence. Geometric sequences are formed by multiplying each term by a constant. To find the nth term in a geometric sequence use the formula: \(a_n=a_1(r^{n−1})\), where \(a_1\) is the first term in the sequence and “n” is the term you’re looking to find.

\(a_n=a_1(r^{n−1})\)

\(a_{600}=2x(r^{n−1})\)

“r” is the ratio, which we can find by dividing the second term by the first term. 4x/(2x) = 2, so r = 2.

\(a_{600}=2x(r^{600−1})\)

\(a_{600}=2x(2^{599})\)

Remember that when we are multiplying by similar bases, we can add the exponents:

\(a_{600}=2^{600}x\)
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Re: In a geometric sequence each term is found by multiplying the previous   [#permalink] 30 Dec 2017, 02:34
Display posts from previous: Sort by

In a geometric sequence each term is found by multiplying the previous

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.