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

It is currently 25 Sep 2018, 06:13

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

There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n

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

Hide Tags

Math Revolution GMAT Instructor
User avatar
V
Joined: 16 Aug 2015
Posts: 6242
GMAT 1: 760 Q51 V42
GPA: 3.82
Premium Member
There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 05 Jun 2016, 06:04
2
7
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (02:32) correct 34% (03:23) wrong based on 86 sessions

HideShow timer Statistics

There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)?
A. 2^22+1
B. 2^23
C. 2^23+1
D. 2^24
E. 2^23+1

*An answer will be posted in 2 days.

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Senior Manager
Senior Manager
User avatar
Joined: 28 Aug 2006
Posts: 302
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 05 Jun 2016, 07:54
Couple of ways to answer this ....

Method 1: Find the pattern

Let's say A(1) = x for a while

Now, let's write the series bearing in mind that each term is 1 less than twice its preceding term.

\(A1 = x\)

\(A2 = 2x -1\)

\(A3 = 4x -3\)

\(A4 = 8x -7\)

\(A5 = 16x -15\)

\(A6 = 32x -31\)
.............
.............

Notice the following

\(A2 - A1 = x - 1\)
\(A3 - A2 = 2(x-1) = 2^1(x-1)\)
\(A4 - A3 = 4(x-1) = 2^2(x-1)\)
\(A5 - A4 = 8(x-1) = 2^3(x-1)\)
\(A6 - A5 = 16(x-1) = 2^4(x-1)\)

\(=> A24 - A23 = 2^{22}(x-1)\)
\(=> A24 - A23 = 2^{22}(3-1) = 2^{23}\) (Given x = 3 in the question)
\(=> A24 - A23 = 2^{23}\)

Method 2: Going to the start of the series

Given \(A(n+1) = 2A(n) - 1\)
\(=> A(n+1) + 1 = 2A(n)\) --------(i)

Bearing this format in mind, let us write the given question the following way

\(A(24) - A(23)\)
\(= A(24) + 1 - A(23) - 1\)
\(= [A(24) +1 ] - [A(23) + 1]\)
\(= 2A(23) - 2A(22)\) ( From (i) )
\(= 2[A(23) - A(22)]\)

Similarly \(A(23) - A(22)\) will equal \(2[A(22) - A(21)]\)
\(\therefore A(24) - A(23) = 2^2[A(22) - A(21)]\)

Continuing this, we get

\(A(24) - A(23) = 2^{22}[A2 - A1]\)
\(\therefore A(24) - A(23) = 2^{22}[5 - 3 ] = 2^{23}\)
_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

Current Student
User avatar
Joined: 18 Oct 2014
Posts: 870
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
GMAT ToolKit User
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 05 Jun 2016, 08:47
2
2
MathRevolution wrote:
There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)?
A. 2^22+1
B. 2^23
C. 2^23+1
D. 2^24
E. 2^23+1

*An answer will be posted in 2 days.


A1= 3
A2= 2*3-1= 5
A3= 2*5-1= 9
A4= 2*9-1= 17

We can notice that there is a squence
A2-A1= 2^1
A3-A2= 2^2
A4-A3= 2^3

Hence A24-A23= 2^23

B is the answer
_________________

I welcome critical analysis of my post!! That will help me reach 700+

Math Revolution GMAT Instructor
User avatar
V
Joined: 16 Aug 2015
Posts: 6242
GMAT 1: 760 Q51 V42
GPA: 3.82
Premium Member
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 06 Jun 2016, 21:55
From A(1)=3=2+1, A(2)=2A(1)-1=5=2^2+1,…A(n)=2^n+1, we can calculate A(24)-A(23)=(2^24+1)-(2^23-1)=2^24-2^23=(2-1)2^23=2^23. Hence, the correct answer is B.

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Manager
Manager
User avatar
S
Joined: 01 Dec 2016
Posts: 113
Concentration: Finance, Entrepreneurship
GMAT 1: 650 Q47 V34
WE: Investment Banking (Investment Banking)
GMAT ToolKit User
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 29 Mar 2017, 00:58
I missed it because i started iteration by A(24) - A(23) and tried to decrease it.

It is better to start with A(2) - A(1), increase interations and infer A(24)-A(23).

I like the question!!!
_________________

What was previously considered impossible is now obvious reality.
In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them

Manager
Manager
avatar
S
Joined: 13 Dec 2013
Posts: 161
Location: United States (NY)
Concentration: Nonprofit, International Business
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)
Reviews Badge
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 12 Apr 2017, 18:19
1
MathRevolution wrote:
There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)?
A. 2^22+1
B. 2^23
C. 2^23+1
D. 2^24
E. 2^23+1

*An answer will be posted in 2 days.



Working out the first few terms gives:

A1=3
A2=5
A3=9
A4=17
A5=33

The differences between each of the terms are 2, 4, 8, 16. Each term increases by 2 to an increased power. A1 increases by 2^1, A2 increases by 2^2 and A3 increases by 2^3 and so on.
So A23 will increase by 2^23 and that's the difference.

Is this the correct method? Is there a better one?
Manager
Manager
avatar
S
Joined: 18 Oct 2016
Posts: 139
Location: India
WE: Engineering (Energy and Utilities)
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 18 Apr 2017, 07:26
Option B

A(n+1) = 2A(n) - 1; A(1) = 3.

A(2) = 2*3 - 1 = 5 = 2^2 + 1
A(3) = 2*5 - 1 = 9 = 2^3 + 1
A(4) = 2*9 - 1 = 17 = 2^4 + 1
A(5) = 2*17 - 1 = 33 = 2^5 + 1
.
.
A(23) = 2^23 + 1
A(24) = 2^24 + 1

A(24) - A(23) = 2^23(2-1) = 2^23
_________________

Press Kudos if you liked the post!

Rules for posting - PLEASE READ BEFORE YOU POST

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8177
Premium Member
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

Show Tags

New post 19 Sep 2018, 22:52
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n &nbs [#permalink] 19 Sep 2018, 22:52
Display posts from previous: Sort by

There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n

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

Events & Promotions

PREV
NEXT


Copyright

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®.