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

It is currently 17 Aug 2018, 13:54

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) where n is a positive integer such that A(n+1

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

Hide Tags

Intern
Intern
avatar
B
Joined: 05 Jun 2016
Posts: 18
GMAT 1: 760 Q51 V41
Reviews Badge
There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post Updated on: 15 Jan 2017, 01:32
4
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

52% (01:57) correct 48% (02:34) wrong based on 66 sessions

HideShow timer Statistics

There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Originally posted by hwang327 on 14 Jan 2017, 14:21.
Last edited by Bunuel on 15 Jan 2017, 01:32, edited 1 time in total.
Renamed the topic and edited the question.
Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 15 Jan 2017, 01:00
2
1
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Any explanations would be great.

Source: MathRevolution


Hi

This is combination of geometric and arithmetic sequences.

n>0, min n = 1.

\(a_2 = 10 + \frac{1}{2}a_1\)

\(a_3 = 10 + \frac{1}{2}a_2 = 10 + \frac{1}{2}(10 + \frac{1}{2}a_1) = 10 + 5 + \frac{1}{4}a_1\)

\(a_4 = 10 + \frac{1}{2}a_3 = 10 + \frac{1}{2}(10 + 5 + \frac{1}{4}a_1) = 10 + 5 + \frac{5}{2} + \frac{1}{8}a_1\)

\(a_5 = 10 + \frac{1}{2}a_4 = 10 + 5 + \frac{5}{2} + \frac{5}{4} + \frac{1}{16}a_1\)

...

\(a_n = 10 + 5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{n-3}} + \frac{1}{2}^{n-1}a_1\)

When n=1000 our fraction \(\frac{1}{2^{999}}\) is close to 0.

\(5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{997}}\) We can apply logic of infinite geometric series where |r|<1 because our n is quite big (1000).

\(S = \frac{5}{1-1/2} = 5*2 = 10\)

\(a_{1000} ≈ 10 + 10 + 0 = 20\)

Answer C

Hope this helps

Regards
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6544
There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 15 Jan 2017, 04:49
2
1
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50



Hi,


A point before the solution..
The Q is flawed in that there is no value of \(A_1\) given.
Solution..
\(A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....\)
So 1000 can be taken as infinite series..
Ans =\(\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20\)
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Intern
Intern
avatar
S
Joined: 30 Jan 2016
Posts: 7
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 16 Jan 2017, 17:03
chetan2u wrote:
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50



Hi,


A point before the solution..
The Q is flawed in that there is no value of \(A_1\) given.
Solution..
\(A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....\)
So 1000 can be taken as infinite series..
Ans =\(\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20\)



Can you please explain to me how you got to the 10/(1-1/2) part in the last equation? I can not seem to trace the origin of the 1/2 part and why that expression is the divisor of 10.
Thank you.
Manager
Manager
User avatar
G
Joined: 22 May 2015
Posts: 106
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 16 Jan 2017, 18:24
Given : A(n+1) = 10 + A(n)/2

A(2) = 10+ A(1)/2
A(3) = 10+ A(2)/2 = 15+A(1)/4
A(4) = 10+ A(3)/2 = 17.5+A(1)/8
A(5) = 10+ A(4)/2 = 18.75+A(1)/16
A(6) = 10+A(5)/2 = 19.375+A(1)/32
a(7) = 19.6875 + a(1)/64
so the second term for A(1000) somewhat 19.XxXXXXXx+A(1)/2^999 , the second part can be ignored closet answer would be 20.
_________________

Consistency is the Key

Intern
Intern
avatar
S
Joined: 30 Jan 2016
Posts: 7
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 19 Jan 2017, 14:58
laddaboy wrote:
Given : A(n+1) = 10 + A(n)/2

A(2) = 10+ A(1)/2
A(3) = 10+ A(2)/2 = 15+A(1)/4
A(4) = 10+ A(3)/2 = 17.5+A(1)/8
A(5) = 10+ A(4)/2 = 18.75+A(1)/16
A(6) = 10+A(5)/2 = 19.375+A(1)/32
a(7) = 19.6875 + a(1)/64
so the second term for A(1000) somewhat 19.XxXXXXXx+A(1)/2^999 , the second part can be ignored closet answer would be 20.


I see now, throughout all equations you utilize A(1) and the end result is neglectable. Thanks a million!
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 7746
Premium Member
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

Show Tags

New post 27 Jul 2018, 01:54
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

Re: There is a sequence A(n) where n is a positive integer such that A(n+1 &nbs [#permalink] 27 Jul 2018, 01:54
Display posts from previous: Sort by

There is a sequence A(n) where n is a positive integer such that A(n+1

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

Events & Promotions

PREV
NEXT


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®.