Veritas Prep 10 Year Anniversary Promo Question #3 : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 06:41

### 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

# Veritas Prep 10 Year Anniversary Promo Question #3

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36582
Followers: 7086

Kudos [?]: 93246 [0], given: 10555

Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:00
Expert's post
5
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:36) correct 32% (02:31) wrong based on 174 sessions

### HideShow timer Statistics

Veritas Prep 10 Year Anniversary Promo Question #3

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course (\$1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

A sequence is given by the rule $$a_{n} = |a_{(n-2)}| - |a_{(n-1)}|$$ for all $$n\geq{3}$$, where $$a_1 = 0$$ and $$a_2 = 3$$.

A function $$s_{n}$$ is defined as the sum of all the terms of the sequence from its beginning through $$a_n$$. For instance, $$s_{4} = a_1 + a_2 + a_3 + a_4$$. What is $$s_{101}$$?

(A) -3
(B) 0
(C) 3
(D) 201
(E) 303
[Reveal] Spoiler: OA

_________________
 Veritas Prep GMAT Discount Codes EMPOWERgmat Discount Codes e-GMAT Discount Codes
Math Expert
Joined: 02 Sep 2009
Posts: 36582
Followers: 7086

Kudos [?]: 93246 [0], given: 10555

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:00
Expert's post
2
This post was
BOOKMARKED

Winner:

yogeshwar007

Official Explanation:

While we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “pre-previous” one.

So $$a_3 = |a_1| - |a_2|$$, and $$a_4 = |a_2| - |a_3|$$, and $$a_5 = |a_3| - |a_4|$$, and so on. We are told that the sequence begins 0, 3, ..., so we can derive that:

$$a_3 = |a_1| - |a_2| = |0| - |3| = 0 - 3 = -3$$.
Then $$a_4 = |a_2| - |a_3| = |3| - |-3| = 3 - 3 = 0$$.
Then $$a_5 = |a_3| - |a_4| = |-3| - |0| = 3 - 0 = 3$$.

As soon as we’ve seen $$a_4$$ and $$a_5$$ turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as $$a_1$$ and $$a_2$$) — so $$a_6$$ will be identical to $$a_3$$, so -3. So we have established the pattern 0, 3, -3, 0, 3, -3, ... for our sequence.

Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + -3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with $$a_{99}$$. $$a_{100}$$ will then add itself (0) to that sum, and $$a_{101}$$ will add itself (3) onto that. So we will land at a sum of 3 for $$s_{101}$$.
_________________
Director
Joined: 24 Aug 2009
Posts: 507
Schools: Harvard, Columbia, Stern, Booth, LSB,
Followers: 17

Kudos [?]: 676 [0], given: 276

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:05
Answer is E as the sum of any first two terms is 3 & then 6, 6, 12, 12 & so on
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Intern
Joined: 21 Oct 2011
Posts: 10
Followers: 0

Kudos [?]: 11 [0], given: 9

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:08
first 8 terms 0,3,-3,0,3,-3,0,3

sum of every four terms is 0+3-3+0 = 0 sum of S100 = 0 sum of s101 = 0+3 =3

Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 647
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Followers: 46

Kudos [?]: 547 [1] , given: 23

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:10
1
KUDOS
Ans is C,

This would be a series of -3 , 0, 3,0 ... .
_________________

Lets Kudos!!!
Black Friday Debrief

Last edited by Vips0000 on 18 Sep 2012, 09:40, edited 1 time in total.
Intern
Joined: 05 Jun 2012
Posts: 22
Followers: 0

Kudos [?]: 0 [0], given: 3

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:12

S101 = a1+ a2+ .....+ a101

a3=|a1|-|a2|= -3
a4=0

Follwoing the pattern,,

0,3,-3,0,3,-3,....

S99 = 0

A100=0

A101=3

Hence,S101=3
Intern
Joined: 05 Sep 2012
Posts: 2
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:14
1
KUDOS

a = {0,3,-3,0,3,-3,0,3,-3....} repeated every 3 counts
s101 = contains 33 triplets + 2 so the sum is 33 * (0+3-3) + (0+3) = 3

Last edited by jirayr on 18 Sep 2012, 09:35, edited 1 time in total.
Manager
Joined: 20 Nov 2010
Posts: 224
Followers: 4

Kudos [?]: 26 [0], given: 38

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:18
IMo - C

pattern
0,3,-3, ...
sum of 3 terms = 0
sum of 99 terms = 0
100 th = 0
101st = 3

sum = 3
_________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
MGMAT 6 650 (51,31) on 31/8/11
MGMAT 1 670 (48,33) on 04/9/11
MGMAT 2 670 (47,34) on 07/9/11
MGMAT 3 680 (47,35) on 18/9/11
GMAT Prep1 680 ( 50, 31) on 10/11/11

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned

Intern
Joined: 09 Aug 2012
Posts: 4
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:47
Ans=C,
i have considered first upto a12, and observed that s1=0, s2=3, s3=0 and the cycle repeats..
by this we can divide 101 as 99+2 i.e., s99 would be 0 and s100 =0 and s101=3.

Last edited by chetanachethu on 18 Sep 2012, 10:05, edited 2 times in total.
BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1196
Followers: 129

Kudos [?]: 1465 [0], given: 142

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 09:51
Considering the rule of Cyclicity,we can conclude that the pattern repeat itself after 3 units i.e a1=0,a2=3,a3=-3 ; again a4=0,a5=3,a6=-3 and so on...

Now 101/3 [as the pattern repeat itself after 3 units as above] gibes a quotient 99 and thus sum of a1+a2+..a99=0 [as every 3 units having sum 0]and a100+a101=|0|+|3|=3

So, S101=a1+a2+..a99+a100+a101=3.

_________________
Intern
Joined: 09 Aug 2012
Posts: 1
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 10:24

We have
an = |a(n-2)| - |a(n-1)|
a1 = 0
a2 =3

If we use the above formula we get a repetitve series of 3,0,-3 as follows
a3 = |a2|- |a1| = 3-0 = 3
a4 = |a3| - |a2| = 3-3 =0
a5 = |a4|-|a3| = 0 - 3 = -3
a6 = |a5| - |a4| = |-3| - 0 = 3-0 =3
a7 = |a6| - |a5| = |3| - |-3| = 3 -3 = 0
And this trend will continue as follows a8 = -3 , a9 = 3 ....and so on until a101 = -3
If we observe there is a pattern emerging starting a3 .If we sum a particular triplet,say a3 + a4+ a5 = 3+0+-3 = 0.
Our sum will be 0
Similiarly if we keep adding triplets
a6 + a7 + a8 =0
a9 + a10 + a11 = 0
.
.
.
a99 + a100 + a101 = 3 + 0 + -3 = 0.
So we can conclude the sum of all the triplets is 0.
a3 + .a4 + a5 + .......a101 = 0.

S101 = a1 + a2 + (sum of all the triplets present )
= 0 + 3 + 0
Hence S101 = 3.

Manager
Joined: 12 Mar 2012
Posts: 172
Location: India
Concentration: Technology, General Management
GMAT Date: 07-23-2012
WE: Programming (Telecommunications)
Followers: 0

Kudos [?]: 51 [0], given: 4

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 10:38
Bunuel wrote:
A sequence is given by the rule $$a_{n} = |a_{(n-2)}| - |a_{(n-1)}|$$ for all $$n\geq{3}$$, where $$a_1 = 0$$ and $$a_2 = 3$$.

A function $$s_{n}$$ is defined as the sum of all the terms of the sequence from its beginning through $$a_n$$. For instance, $$s_{4} = a_1 + a_2 + a_3 + a_4$$. What is $$s_{101}$$?

(A) -3
(B) 0
(C) 3
(D) 201
(E) 303

a3 = |a1| - |a2|
a4 = |a2| - |a3|
a5 = |a3| - |a4|
..
..
..
..
a100 = |a98| - |a99|
a101 = |a99| - |a100|
----------------------
Summing together both sides
a3+a4+a5+.......+a101 = |a1| - |a100| ( because all other values will be cancelled out in right hand side. For example, |a2| from 1st equation will be cancelled out from |a2| from 2nd equation, and so on.)

adding a1 + a2 both sides
a1 + a2 +a3 .............. + a101 = S101 = a1+a2+|a1| - |a100|
=> 0 + 3 + |0| - |a100|

Now these terms follow a pattern as follows:
a3 = -3
a4 = 0
a5 = -3
..
..
..
So, EVEN indexed "a" will be zero.
Thus, => 0+3+|0|+0 = 3

_________________

FOCUS..this is all I need!

Ku-Do!

Intern
Joined: 30 Mar 2012
Posts: 6
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 10:56
a1=0, a2=3, a3=-3, a4=0, a5=3, a6=-3.......
repetition occurs after every 3rd unit/count.
S101=a1+a2+a3....+a99+a100+a101
s101=0+a100+a101 as [101][/3] leaves a remainder 2; therefore sum (a1+a2+a3+...a99)=0
S101=0+0+3
=3, (c)
Intern
Joined: 08 Sep 2012
Posts: 3
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 12:08
So I figured out the sequence first
a1=0, a2=3, a3=-3, a4=0, a5=3, a6=-3, a7=0, a8=3, a9=-3, a10=0.......
Every third term is a -3 for a3, a6, a9, a12, a15, a18, a21, a, 24....... They are all -3.
So s102 = -3 therefore s101 = 3
Another is that the sum of the first 10 is 0 s10= s1 + s2 +s3 + s4....... is 0 so every tenth interval is 0, therefore s100=0
The answer is 3 which is option C
Manager
Joined: 14 Feb 2012
Posts: 55
Location: United States
WE: Project Management (Consulting)
Followers: 1

Kudos [?]: 18 [0], given: 0

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 12:46
A sequence is given by the rule $$a_{n} = |a_{(n-2)}| - |a_{(n-1)}|$$ for all $$n\geq{3}$$, where $$a_1 = 0$$ and $$a_2 = 3$$.

A function $$s_{n}$$ is defined as the sum of all the terms of the sequence from its beginning through $$a_n$$. For instance, $$s_{4} = a_1 + a_2 + a_3 + a_4$$. What is $$s_{101}$$?[/b]

(A) -3
(B) 0
(C) 3
(D) 201
(E) 303

Explaination:
a(1) = 0 (given)
a(2) = 3 (given)
a(3) = |a(1)| - |a(2)| = |0| - |3| = 0 - 3 = -3
a(4) = |a(2)| - |a(3)| = |3| - |-3| = 3 - 3 = 0
a(5) = |a(3)| - |a(4)| = |-3| - |0| = 3 - 0 = 3
a(6) = |a(4)| - |a(5)| = |0| - |3| = 0 - 3 = -3

So the series is 0, 3, -3, 0, 3, -3, for a(1), a(2), a(3), a(4), a(5) and a(6) respectively. If you look at the series carefully, S(3) = 0 + 3 + (-3) = 0
this trend will continue and hence S(3), S(6), S(9)..... S(99) = 0
S(101) = S(99) + a(100) + a(101) = 0 + 0 + 3 = 3
S(101) = 3

Current Student
Joined: 07 Sep 2011
Posts: 74
GMAT 1: 660 Q41 V40
GMAT 2: 720 Q49 V39
WE: Analyst (Mutual Funds and Brokerage)
Followers: 3

Kudos [?]: 45 [0], given: 13

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 13:06

The pattern is 0,3,-3,0,3,-3,0,3,-3...

101/3=99 R2

0+3=3
Intern
Joined: 03 Jul 2012
Posts: 5
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 17:12

Solution:
We have: S101 =a1+ a2+a3+a4…..+a99+a100+a101
= a1+ a2+|a1 |-|a2 |+|a2 |-|a3 |+⋯.+|a97 |-|a98 |+|a98 |-|a99 |+|a99 |-|a100 |
= a1+ a2+|a1 |-|a100 |
= 3-|a100 | (1)

Then: a1=0; a2=3; a3=|a1 |-|a2 |=0-3=-3; a4=|a2 |-|a3 |=3-3=0; a5=|a3 |-|a4 |=3-0=3;
Similarly: a4=0; a5=3; a6=|a4 |-|a5 |=0-3=-3; a7=|a_5 |-|a6 |=3-3=0; a8=|a6 |-|a7 |=3-0=3;
From those results,we can conclude that: a(3n+1)=0;a(3n+2)=3;a(3n)=-3

So, a_100=a(3×33+1)=0 (2)
From (1) and (2), we have: S101= 3-|a100 | = 3
Intern
Joined: 25 Jul 2012
Posts: 1
Followers: 0

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

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 18:29

Explaination:
a(1) = 0 (given)
a(2) = 3 (given)
a(3) = |a(1)| - |a(2)| = |0| - |3| = 0 - 3 = -3
a(4) = |a(2)| - |a(3)| = |3| - |-3| = 3 - 3 = 0
a(5) = |a(3)| - |a(4)| = |-3| - |0| = 3 - 0 = 3
a(6) = |a(4)| - |a(5)| = |0| - |3| = 0 - 3 = -3

However the question asks for S(101). So we have to find a series for S(n).
S(1) = a(1) = 0
S(2) = a(1) + a(2) = 0 + 3 = 3
S(3) = a(1) + a(2) + a(3) = 0 + 3 - 3 = 0
S(4) = a(1) + a(2) + a(3) + a(4) = 0 + 3 - 3 + 0 = 0
s(5) = a(1) + a(2) + a(3) + a(4) + a(5) = 0 + 3 - 3 + 0 + 3 = 3
s(6) = a(1) + a(2) + a(3) + a(4) + a(5) +a(6) = 0 + 3 - 3 + 0 + 3 - 3 = 0

So the real series is 0, 3, 0, 0, 3, 0, for s(1), s(2), s(3), s(4), s(5) and s(6) respectively.

If you divide 101/3 you get 33 remainder 2, which tells us the correct answer should be the 2nd term of the series,
hence S(101) = 3
Intern
Status: Winning is not everything, but wanting to win is
Joined: 10 Nov 2010
Posts: 35
WE 1: IT Consultant (6Yr)
Followers: 0

Kudos [?]: 9 [0], given: 5

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

18 Sep 2012, 21:23

Solution

As per the question we have values of terms
a1 = 0;a2 =3;a3 = -3

Using the function $$a_{n} = |a_{(n-2)}| - |a_{(n-1)}|$$ for all $$n\geq{3}$$, can derive

a4 = 0

If we continue solving the next two terms (i.e a5 & a6) we will see that these terms are cyclic i.e (0,3,-3)

Moving forward we have,

$$s_{4} = a_1 + a_2 + a_3 + a_4$$

Calculating the first four terms s (4) = 0 ; In the same way s(100) = 0 and s(101) = s(100) + a101 = 3

Hence OA is C
Math Expert
Joined: 02 Sep 2009
Posts: 36582
Followers: 7086

Kudos [?]: 93246 [0], given: 10555

Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink]

### Show Tags

19 Sep 2012, 08:01

We have a winner!

Winner:

yogeshwar007

Official Explanation:

While we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “pre-previous” one.

So $$a_3 = |a_1| - |a_2|$$, and $$a_4 = |a_2| - |a_3|$$, and $$a_5 = |a_3| - |a_4|$$, and so on. We are told that the sequence begins 0, 3, ..., so we can derive that:

$$a_3 = |a_1| - |a_2| = |0| - |3| = 0 - 3 = -3$$.
Then $$a_4 = |a_2| - |a_3| = |3| - |-3| = 3 - 3 = 0$$.
Then $$a_5 = |a_3| - |a_4| = |-3| - |0| = 3 - 0 = 3$$.

As soon as we’ve seen $$a_4$$ and $$a_5$$ turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as $$a_1$$ and $$a_2$$) — so $$a_6$$ will be identical to $$a_3$$, so -3. So we have established the pattern 0, 3, -3, 0, 3, -3, ... for our sequence.

Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + -3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with $$a_{99}$$. $$a_{100}$$ will then add itself (0) to that sum, and $$a_{101}$$ will add itself (3) onto that. So we will land at a sum of 3 for $$s_{101}$$.
_________________
Re: Veritas Prep 10 Year Anniversary Promo Question #3   [#permalink] 19 Sep 2012, 08:01

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
15 The age of the Earth is approximately 1.3*10^17 seconds, and one year 6 28 Oct 2014, 16:37
5 The distance between two planets is 3.04 × 10^6 light years. 3 16 Oct 2013, 11:23
8 Veritas Prep 10 Year Anniversary Promo Question #9 24 21 Sep 2012, 09:00
2 Veritas Prep 10 Year Anniversary Promo Question #1 18 17 Sep 2012, 09:00
Spaceland Prep Strategy Question #3 3 10 Mar 2011, 07:50
Display posts from previous: Sort by