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

 It is currently 18 Jan 2019, 22:17

### 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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
• ### FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# In a certain sequence, each term, starting with the 3rd term

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

### Hide Tags

Manager
Status: struggling with GMAT
Joined: 06 Dec 2012
Posts: 123
Location: Bangladesh
Concentration: Accounting
GMAT Date: 04-06-2013
GPA: 3.65
In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

27 Mar 2013, 21:53
6
18
00:00

Difficulty:

95% (hard)

Question Stats:

45% (02:18) correct 55% (02:27) wrong based on 819 sessions

### HideShow timer Statistics

In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?

(1) The 1st term is equal to 8 times the 2nd term.
(2) The 4th term is equal to 1.
##### Most Helpful Community Reply
Retired Moderator
Joined: 15 Jun 2012
Posts: 1011
Location: United States
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

27 Jun 2013, 22:18
6
1
mun23 wrote:
In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?

(1) The 1st term is equal to 8 times the 2nd term.
(2) The 4th term is equal to 1.

The question is:
1st term = $$x$$
2nd term = $$y$$
3rd term = $$xy$$
4th term = $$(xy)*y = xy^2$$
5th term = $$(xy^2)*(xy) = x^2y^3$$
6th term = $$(x^2y^3)*(xy^2) = x^3y^5$$

Have equation: 6th term - 3rd term $$= x^3y^5 - xy = xy(x^2y^4 -1)$$

Statement 1: The 1st term is equal to 8 times the 2nd term ==> $$x = 8y$$
Replace back to equation: 6th term - 3rd term $$= 8yy((8y)^2y^4 -1) = 8y^2(64y^6 - 1)$$ ==> NOT Sufficient because we don't know y

Statement 2: The 4th term is equal to 1 ==> $$=xy^2 = 1$$.
Replace back to equation: 6th term - 3rd term $$= xy((xy^2)^2 - 1) = xy(1 - 1) = 0$$ ==> Sufficient

Hence, B is correct.

Hope it helps.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Chris Bangle - Former BMW Chief of Design.

##### General Discussion
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 611
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

27 Mar 2013, 22:40
4
2
mun23 wrote:
In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?

(1) The 1st term is equal to 8 times the 2nd term.

(2) The 4th term is equal to 1.

Need explanation

From F.S 1, consider this series : 8, 1, 8, 8, 64, 64*8.

Again, consider the series 2, 1/4, 1/2, 1/8, 1/16, 1/(8*16). Clearly, the difference of the 6th and the 3rd term is different for them. Insufficient.

From F.S 2, let the series be $$a,b,ab,ab^2,a^2b^3,a^3b^5$$. Now we know that $$ab^2 = 1$$. The required difference =$$a^3b^5 - ab = ab(a^2b^4-1) = ab[(ab^2)^2 -1]$$= 0.Sufficient.

B.
_________________
Manager
Joined: 21 Oct 2013
Posts: 185
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

04 Feb 2014, 04:04
(1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.

(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.

Hence answer B.

I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds!
SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

18 Feb 2014, 16:29
unceldolan wrote:
(1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.

(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.

Hence answer B.

I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds!

Why should all terms be 1? Do we have to assume that all terms are integers here? What if S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1

Can anyone explain second statement?

Thanks a lot
Cheers
J
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8792
Location: Pune, India
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

18 Feb 2014, 21:13
2
1
jlgdr wrote:
unceldolan wrote:
(1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.

(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.

Hence answer B.

I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds!

Why should all terms be 1? Do we have to assume that all terms are integers here? What if S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1

Can anyone explain second statement?

Thanks a lot
Cheers
J

No, you don't have to assume anything. What you say is absolutely possible though it doesn't change our answer.

"S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1"

The sequence could be: 9, 1/3, 3, 1, 3, 3, ....
But in any case $$t_6 - t_3 = 0$$

Note why: Statement 2 tells us that $$t_4 = 1$$

$$t_1, t_2, t_3, 1,$$
what will be $$t_5$$? It will be $$t_3 * 1 = t_3$$
What will be $$t_6$$? It will be $$1*t_5 = 1*t_3 = t_3$$

So in any case the sixth term will be same as the third term. Once you have a 1 in the sequence, all following terms will be equal to the term just preceding 1.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Manager
Joined: 18 May 2014
Posts: 58
Location: United States
Concentration: General Management, Other
GMAT Date: 07-31-2014
GPA: 3.99
WE: Analyst (Consulting)
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

18 May 2014, 09:39
1
The answer is B...

First Term to Sixth
1..-1..-1...1...-1...1

Diff b/w 6th and 3rd is -1 -(-1)=0
_______________________
1/64...8...1/8...1...1/8....1/8...1/64...

Diff b/w 1/64-1/8=-7/64
------------------------
SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

27 May 2014, 05:33
2
Shouldn't be hard at all, let's see

So we need to find S6 - S3

Now, S6 = (S5)(S4)

So (S5)(S4) - S3?

Statement 1

Clearly insufficient. We don't have much information here as given that S=8(S2)

Statement 2

Now, here's where it gets interesting. We have a specific value S4=1.

Now, since S4=1 we are being asked what is S5-S3?

S5 = (S4)*(S3). Again since S4=1 then S5=S3.

So S5-S3=0

Sufficient

Answer: B

Hope this helps!
Cheers
J
Intern
Joined: 15 Sep 2013
Posts: 32
Concentration: Strategy, Entrepreneurship
GMAT 1: 680 Q47 V36
GMAT 2: 740 Q50 V40
GPA: 3.65
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

04 Aug 2014, 09:23
mun23 wrote:
In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?

(1) The 1st term is equal to 8 times the 2nd term.
(2) The 4th term is equal to 1.

Statement 1 is clearly insufficient as it tells nothing about 4th or 5th term
Statement 2 says that the 4th term is 1
Let the 3rd term be anything.... lets say the 3rd term is 'a' and the 4th term is 1
Thus 5th term would be a*1.. which is equal to a
And 6th term would be again 5th term*4th term that is a*1=a
Thus no matter what the 3rd term be.... if the 4th term is 1 then the 3rd and 6th term would always be equal..
thus the difference would be 0 - Sufficient.
_________________

Please +1 KUDOS if my post helped you in any way

Manager
Joined: 14 Jul 2014
Posts: 170
Location: United States
Schools: Duke '20 (D)
GMAT 1: 600 Q48 V27
GMAT 2: 720 Q50 V37
GPA: 3.2
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

29 Dec 2014, 19:09
Could someone explain why the 3rd term is not 1 when the 4th is 1? Doesn't the 4th term have to be the product of the 2nd and the 3rd term?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8792
Location: Pune, India
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

29 Dec 2014, 19:35
1
dina98 wrote:
Could someone explain why the 3rd term is not 1 when the 4th is 1? Doesn't the 4th term have to be the product of the 2nd and the 3rd term?

$$T_4 = T_3 * T_2 = 1$$
$$T_3$$ could be anything and $$T_2$$ would take a corresponding value.
For example, $$T_3$$ could be 2 and $$T_2$$ would be 1/2. Their product would be 1. Their product needs to be 1, they both independently don't need to be 1.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Manager
Joined: 22 Feb 2016
Posts: 90
Location: India
Concentration: Economics, Healthcare
GMAT 1: 690 Q42 V47
GMAT 2: 710 Q47 V39
GPA: 3.57
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

24 Nov 2016, 19:48
1
pqhai wrote:
mun23 wrote:
In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?

(1) The 1st term is equal to 8 times the 2nd term.
(2) The 4th term is equal to 1.

The question is:
1st term = $$x$$
2nd term = $$y$$
3rd term = $$xy$$
4th term = $$(xy)*y = xy^2$$
5th term = $$(xy^2)*(xy) = x^2y^3$$
6th term = $$(x^2y^3)*(xy^2) = x^3y^5$$

Have equation: 6th term - 3rd term $$= x^3y^5 - xy = xy(x^2y^4 -1)$$

Statement 1: The 1st term is equal to 8 times the 2nd term ==> $$x = 8y$$
Replace back to equation: 6th term - 3rd term $$= 8yy((8y)^2y^4 -1) = 8y^2(64y^6 - 1)$$ ==> NOT Sufficient because we don't know y

Statement 2: The 4th term is equal to 1 ==> $$=xy^2 = 1$$.
Replace back to equation: 6th term - 3rd term $$= xy((xy^2)^2 - 1) = xy(1 - 1) = 0$$ ==> Sufficient

Hence, B is correct.

Hope it helps.

Thanks, I did it exactly the same way .
It encourages me to know that I think in the proper direction under stressed situations.
Intern
Joined: 14 May 2014
Posts: 16
In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

24 Nov 2016, 20:56
From (1): a1= 8x, a2=x => a3=8x^2 and a6=8^3*x^8 => not sufficient
From (2): a1 a2 a3 a4(=1) => a5=a3 and a6=1*a5=a3 => the difference between a3 and a6 = 0
==> B is the answer
Manager
Joined: 20 Jan 2016
Posts: 184
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

31 Oct 2017, 10:30
OE

We have been asked for the difference between the 6th and 3rd terms of a sequence. There is little we can do to rephrase the question, but it is important to note that we don’t necessarily need to know the exact values of these terms to find their difference. We can solve as long as we know the relative values of the two terms, i.e. how much larger or smaller the 6th term is than the 3rd.

(1) INSUFFICIENT: We have been given the relationship between the 1st and 2nd terms. If we let x be the value of the 2nd term, then the 1st term is 8x. Since each term is determined by multiplying the previous two terms, we can build the whole sequence in terms of x. We can begin by multiplying 8x, the1st term, and x, the 2nd term, to get the 3rd term, 8x2. If we continue to build, we get the following sequence (starting from the 1st term): 8x, x, 8x2, 8x3, 82x5, 83x8

The difference between the 6th and 3rd terms is 83x8 − 8x2, but we can’t calculate the actual value without knowing the value of x.

(2) SUFFICIENT: At first glance, this statement does not appear to provide much helpful information. However, let’s look at what this means for the rest of the sequence. The 4th term is 1. We can find the 5th term by multiplying the 3rd and 4th terms. Let’s use n to represent the unknown 3rd term, and see what happens.
3rd term: n
4th term: 1
5th term: n*1 = n
6th term: 1*n = n

The presence of the 1 in the 4th spot produces a pair of “copycat” terms afterward! In other words, the 6th and 3rd terms are equal. Therefore, the difference between them is 0 and the correct answer is B.

Alternatively, we could approach this problem using Smart Numbers.

(1) INSUFFICIENT: We can try different numbers for the first two terms to see whether we get different results.

If we make the first two terms 8 and 1, we produce this sequence:
8, 1, 8, 8, 82, 83
What other numbers can we try? If we increase our 2nd term by 1, we get 16 and 2, but from there we will swiftly find ourselves multiplying some very large numbers! Fortunately, we are free to adjust downward and use fractions. Let’s try 4 and 1/2:
4, ½, 2, 1, 2, 2
Since 83 − 8 is clearly not the same value as 2 − 2, statement 1 is not sufficient.

(2) SUFFICIENT: While we can now use a number rather than a variable to represent the 3rd term, the result will look a lot like our previous solution:
3rd term: 4
4th term: 1
5th term: 4*1 = 4
6th term: 1*4 = 4

No matter what number we plug in, the 3rd and 6th terms must be equal, so their difference will always be 0.

The correct answer is B.
Senior Manager
Joined: 29 Jun 2017
Posts: 464
GPA: 4
WE: Engineering (Transportation)
Re: In a certain sequence, each term, starting with the 3rd term  [#permalink]

### Show Tags

08 Jan 2018, 06:17
1
1) a1,a2,a3,a4,a5,a6
8a,a,8a^2,8a^3,64a^5,512a^8

no info about a
INSUFF

2) a4=1
a3=a3
a4=1
a5=a3
a6=a3
a6-a3 =0

SUFFICIENT B
_________________

Give Kudos for correct answer and/or if you like the solution.

Re: In a certain sequence, each term, starting with the 3rd term &nbs [#permalink] 08 Jan 2018, 06:17
Display posts from previous: Sort by

# In a certain sequence, each term, starting with the 3rd term

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

 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®.