Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 19 Jul 2019, 09:35

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

Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co

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

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56300
Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co  [#permalink]

Show Tags

New post 17 Jun 2019, 01:40
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (03:00) correct 32% (03:21) wrong based on 28 sessions

HideShow timer Statistics


Manager
Manager
avatar
B
Joined: 27 Mar 2016
Posts: 92
GMAT ToolKit User
Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co  [#permalink]

Show Tags

New post 17 Jun 2019, 01:59
2
23,27,28,32,33,37..we can see that each pair sums up to be 50, 60, 70 and so on till 130 as there 9 pairs
Hence sums up to be 810 till 18th term
Since we know each pair has a difference of 4 therefore 17th and 18th term = 2x+4 =130
17 th term = 63
Hence 19th term = 63+5 =68
Hence sum = 810+68=878

Posted from my mobile device
Director
Director
avatar
G
Joined: 20 Jul 2017
Posts: 547
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co  [#permalink]

Show Tags

New post 17 Jun 2019, 02:04
Bunuel wrote:
Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It continues in such a way that by adding 5 to the nth term, one obtains the (n + 2)th term. What is the sum of the first 19 members of this sequence?

A. 878
B. 900
C. 788
D. 928
E. 1022



Let first number = 23
Sequence is 23, 23+4, 23+5, 23+9, 23+10, . . . .
In first 19, there are 10 Odd terms & 9 Even terms
Odd terms \(t_1\), \(t_3\), \(t_5\). . . . . . . .\(t_{19}\) are 23, 23+5, 23+10 . . . . . . 23+45 --> AP series with common difference 5
Even terms \(t_2\), \(t_4\), \(t_6\). . . . . . . .\(t_{18}\) are 23+4, 23+9, 23+14, . . . . . . 23+44 --> AP series with common difference 5

Sum of Odd terms = n/2*(1st term + last term) = 10/2*(23 + 68) = 455
Sum of Even terms =9/2*(1st term + last term) = 9/2*(27 + 67) = 9/2*94 = 423
Total sum = 455 + 423 = 878

IMO Option A

Pls Hit Kudos if you like the solution
Intern
Intern
User avatar
B
Status: Focus - Determination - Domination
Joined: 02 Mar 2015
Posts: 39
Location: India
Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co  [#permalink]

Show Tags

New post Updated on: 17 Jun 2019, 21:55
This problem consists of two series for
1st Series starting with 23,28.33......10 terms
2nd Series starting with 27,32....9 terms

For 1st series,
total number of terms will be 10
and 1stterm =23 & c.d=5
S1=455

Second series will be of 9 terms,
1st term =27 & c.d=5
S2=423

Sum of series = 423+455=878
_________________
Let's Rock!

Originally posted by thembaseeker on 17 Jun 2019, 11:46.
Last edited by thembaseeker on 17 Jun 2019, 21:55, edited 1 time in total.
Manager
Manager
avatar
B
Joined: 23 Jul 2015
Posts: 57
Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co  [#permalink]

Show Tags

New post 17 Jun 2019, 18:55
ashudhall wrote:
23,27,28,32,33,37..we can see that each pair sums up to be 50, 60, 70 and so on till 130 as there 9 pairs
Hence sums up to be 810 till 18th term
Since we know each pair has a difference of 4 therefore 17th and 18th term = 2x+4 =130
17 th term = 63
Hence 19th term = 63+5 =68
Hence sum = 810+68=878

Posted from my mobile device


Hey! This is a really easy solution. My only question is that why is it 2x+4=130 and not 2x+5=130 since the stem mentions that the diff between the alternate numbers is 5?
GMAT Club Bot
Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co   [#permalink] 17 Jun 2019, 18:55
Display posts from previous: Sort by

Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It co

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





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