It is currently 22 Jan 2018, 10:04

### 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

# Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 13 Apr 2013
Posts: 259
Location: India
Schools: ISB '19
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]

### Show Tags

17 Sep 2016, 09:46
5
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

51% (02:53) correct 49% (02:17) wrong based on 69 sessions

### HideShow timer Statistics

Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15
[Reveal] Spoiler: OA

_________________

"Success is not as glamorous as people tell you. It's a lot of hours spent in the darkness."

Senior Manager
Joined: 23 Apr 2015
Posts: 331
Location: United States
WE: Engineering (Consulting)
Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]

### Show Tags

17 Sep 2016, 19:32
GMATantidote wrote:
Sequence $$P$$ is defined by $$p_n$$ = $$p_{n-1}$$ $$+ 3$$, $$p_1$$ $$= 11$$, sequence $$Q$$ is defined as $$q_n$$ = $$q_{n-1}$$ $$– 4$$, $$q_3$$ $$= 103$$. If $$p_k$$ > $$q_{k+2}$$, what is the smallest value $$k$$ can take?

A. 6
B. 9
C. 11
D. 14
E. 15

Set P = {11, 14, 17,.....}
Set Q = {110, 107, 103,....}

For Arithmetic progression $$a(n) = a(1) + (n-1) d$$
Started with C) 11 does not satisfy $$p_k$$ > $$q_{k+2}$$
and D) 14, $$p_k$$ = $$q_{k+2}$$ So answer is E, no need to calculate.
Intern
Joined: 07 Jun 2016
Posts: 46
GPA: 3.8
WE: Supply Chain Management (Manufacturing)
Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]

### Show Tags

12 Oct 2016, 18:24
GMATantidote wrote:
Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15

I am really struggling trying to obtain the correct answer. I can't seem to set up the math even with the GMAT Math Book. Could someone possibly detail it out, nothing too in-depth but enough to show why 15 is correct? I have Pn = a1 + (n-1)d but don't know how to apply the darn formula...

Manager
Joined: 29 Aug 2008
Posts: 111
Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]

### Show Tags

16 Oct 2016, 11:11
mikeyg51 wrote:
GMATantidote wrote:
Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15

I am really struggling trying to obtain the correct answer. I can't seem to set up the math even with the GMAT Math Book. Could someone possibly detail it out, nothing too in-depth but enough to show why 15 is correct? I have Pn = a1 + (n-1)d but don't know how to apply the darn formula...

Sequence P has a common difference of 3 and is (11, 14, 17......)

Sequence Q has a common difference of -4 and is (111, 107, 103.....)

Nth term of any arithmetic sequence can be written as a(n) = a1 + (n-1)d where a1 is first term and d is common difference

Taking values from options starting with C k = 11

That gives P(11) = P1 + (11-1)3 = 11 + 10*3 = 41

Q(13) = Q1 + (13-1)*(-4) = 111 + 12*(-4) = 111 - 48 = 63

P(11) < Q(13) so it doesn't help.

We need to try the same with 14.

P(14) = P1 + (14-1)*3 = 11 + 13*3 = 11 + 39 = 50

Q(16) = Q1 + (16-1)*(-4) = 111 + 15*(-4) = 111 - 60 = 51

Still P(14) < Q(16) so this can't be the value of K.

You can either solve for K = 15 and check or mark E as the answer as that's the only value greater than 14.

HTH
Senior Manager
Joined: 02 Apr 2014
Posts: 349
Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]

### Show Tags

21 Dec 2017, 21:47
1
KUDOS
algebraic approach:
Q3 = 103 => Q1 = 111

Pn : 11 + (n - 1)3 , n >1
Qn : 111 - (n - 1)4, n > 1

to find min value of k such that Pk > Q(k+2)
11 + (k-1) * 3 > 111 - (k+2-1)*4
=> 11 + 3k - 3 > 111 - 4k - 4
=> 7k > 99 => k > 14 => Answer (E), k = 15

Kudos if you like my approach, need to unlock my Gmat Club Tests
Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined   [#permalink] 21 Dec 2017, 21:47
Display posts from previous: Sort by