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

 It is currently 14 Oct 2019, 01:14 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1

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

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58257
A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags 00:00

Difficulty:   45% (medium)

Question Stats: 73% (02:42) correct 28% (02:55) wrong based on 80 sessions

HideShow timer Statistics

A sequence $$p_1$$, $$p_2$$, $$p_3$$, ... $$p_n$$ is such that $$p_1 = 3$$ and $$p_{n+1} = 2p_n − 1$$ for n ≥ 1, then $$p_{10} − p_9 =$$

(A) $$2^9$$

(B) $$2^{10} − 1$$

(C) $$2^{10}$$

(D) $$2^{11} − 1$$

(E) $$2^{11}$$

_________________
Manager  S
Joined: 17 Jul 2017
Posts: 185
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

So p2=2p1-1=2*3-1=5

p3=2p2-1=2*5-1=9

so series will be like

3,5,9,17,33.....

so 5-3=2=2^1
9-5=4=2^2
17-9=8=2^3

so 2nd term -1st term=2^1
3rd term-2nd term=2^2
like wise
10th term-9th term=2^9

so A is answer
Math Expert V
Joined: 02 Aug 2009
Posts: 7952
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

Bunuel wrote:
A sequence $$p_1$$, $$p_2$$, $$p_3$$, ... $$p_n$$ is such that $$p_1 = 3$$ and $$p_{n+1} = 2p_n − 1$$ for n ≥ 1, then $$p_{10} − p_9 =$$

(A) $$2^9$$

(B) $$2^{10} − 1$$

(C) $$2^{10}$$

(D) $$2^{11} − 1$$

(E) $$2^{11}$$

choices should tell you that the answer has to be in terms of $$2^n$$.
keeping this in mind we can work forward..

$$p_{n+1} = 2p_n − 1...........p_{n+1} = p_n+p_n − 1.............p_{n+1} -p_n=p_n − 1$$ so $$p_{10} − p_9 =p_9-1$$
so let us find $$p_9$$

now $$p_1=3=2^1+1$$..
$$p_2=2*p_1-1=2*3-1=5=2^2+1$$..
$$p_3=2*p_2-1=2*5-1=9=2^3+1$$
thus we get term for nth term
$$p_n=2^n+1$$ and $$p_9=2^9+1$$

and $$p_{10} − p_9 =p_9-1=2^9-1+1=2^9$$

A
_________________
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2974
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

Bunuel wrote:
A sequence $$p_1$$, $$p_2$$, $$p_3$$, ... $$p_n$$ is such that $$p_1 = 3$$ and $$p_{n+1} = 2p_n − 1$$ for n ≥ 1, then $$p_{10} − p_9 =$$

(A) $$2^9$$

(B) $$2^{10} − 1$$

(C) $$2^{10}$$

(D) $$2^{11} − 1$$

(E) $$2^{11}$$

To answer such a question one should always write a few terms (as many as you can calculate till 4th or 5th step) and then follow the patterns that question seeks at 10th term

$$p_1 = 3$$
$$p_2 = 2*3-1 = 5$$ i.e. $$p_2 − p_1 =5-3 = 2^1$$
$$p_3 = 2*5- 1 = 9$$ i.e. $$p_3 − p_2 =9-5 = 2^2$$
$$p_4 = 2*8- 1 = 17$$ i.e. $$p_4 − p_3 =17-9 = 2^3$$
$$p_5 = 2*17- 1 = 33$$ i.e. $$p_5 − p_4 =33-17 = 2^4$$
.....
.....
..... Following the same pattern we can deduce that $$p_{10} − p_9 = 2^9$$
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Senior Manager  S
Joined: 12 Sep 2017
Posts: 301
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

Hello everyone!

How do we know that Pn equals P1?

Kind regards!
Senior Manager  S
Joined: 12 Sep 2017
Posts: 301
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

Hello everyone!

Could someone please explain to me how to pick up the right value?

$$p_{n+1} = 2p_n − 1$$ for n ≥ 1

How to be sure to which value they are referring with :

$$p_{n+1} = 2p_n − 1$$

P1 = 3
P2 = 2(3) - 1 = 5

But, I got confused then I had to pick up the value of Pn, is Pn the result of the whole term or is just the number of the term?

Pn can be either 3 or 1?

Can someone help me please?
Manager  S
Joined: 22 Sep 2018
Posts: 240
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

Show Tags

GMATinsight wrote:
Bunuel wrote:
A sequence $$p_1$$, $$p_2$$, $$p_3$$, ... $$p_n$$ is such that $$p_1 = 3$$ and $$p_{n+1} = 2p_n − 1$$ for n ≥ 1, then $$p_{10} − p_9 =$$

(A) $$2^9$$

(B) $$2^{10} − 1$$

(C) $$2^{10}$$

(D) $$2^{11} − 1$$

(E) $$2^{11}$$

To answer such a question one should always write a few terms (as many as you can calculate till 4th or 5th step) and then follow the patterns that question seeks at 10th term

$$p_1 = 3$$
$$p_2 = 2*3-1 = 5$$ i.e. $$p_2 − p_1 =5-3 = 2^1$$
$$p_3 = 2*5- 1 = 9$$ i.e. $$p_3 − p_2 =9-5 = 2^2$$
$$p_4 = 2*8- 1 = 17$$ i.e. $$p_4 − p_3 =17-9 = 2^3$$
$$p_5 = 2*17- 1 = 33$$ i.e. $$p_5 − p_4 =33-17 = 2^4$$
.....
.....
..... Following the same pattern we can deduce that $$p_{10} − p_9 = 2^9$$

Hi, is there a trick to know what equation to find the pattern? I would not have immediately tried to subtract the term with the proceeding term, and I do not think I have the time to brainstorm on the test to find a pattern like you described above. Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1   [#permalink] 06 Feb 2019, 19:33
Display posts from previous: Sort by

A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1

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

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