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# In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X

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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  31 Mar 2010, 10:47
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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) where x is a positive integer constant. For what value of n is the ratio of An to x(1+x(1+x(1+x(1+x)))) equal to X^5?

A. 8
B. 7
C. 6
D. 5
E. 4
[Reveal] Spoiler: OA
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  31 Mar 2010, 11:33
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ksharma12 wrote:
18. In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3)
where x is a positive integer constant. For what value of n is the ratio of An to
x(1+x(1+x(1+x(1+x)))) equal to X^5?

(A) 8

(B) 7

(C) 6

(D) 5

(E) 4

note: An= A sub n

Can you explain this in detail? I tried expanding out the bottom equation and solving for X to equal x^5. Didnt really work out...

x^5=\frac{x^{(n-1)}+x^n+x^{(n+1)}+x^{(n+2)}+x^{(n+3)}}{x(1+x(1+x(1+x(1+x))))}

--> x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}+x^n+x^{(n+1)}+x^{(n+2)}+x^{(n+3)}

--> take x^{(n-1)} out of the brackets

--> x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x+x^2+x^3+x^4)

--> x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x+x^2+x^3))

--> x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x(1+x+x^2)))

--> x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x(1+x(1+x))))

--> x^6=x^{(n-1)} --> n-1=6 --> n=7

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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  02 Feb 2011, 20:37
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craky wrote:
Its not so hard when you realize how can you solve it, but until that, you spent half of your life.

Oh no you don't. Work smart!

An = x^{n-1} + x^n + x^{n+1} + x^{n+2} + x^{n+3}
e.g. A2 = x + x^2 + x^3 + x^4 + x^5
Notice you can only take x common out of all these terms i.e. the smallest term x^{n - 1}

If \frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5, it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den. Ignore it.
From An, you will be able to take out x^6 common so that \frac{x^6}{x} gives you x^5
So smallest term must be x^6 i.e. x^{n-1}. Therefore, n = 7.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5255 Location: Pune, India Followers: 1271 Kudos [?]: 6302 [4] , given: 173 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 04 Feb 2011, 07:20 4 This post received KUDOS Expert's post subhashghosh wrote: Hi Karishma What is the meaning of this ? "it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den." Regards, Subhash A2 = x + x^2 + x^3 + x^4 + x^5 \frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5 Since the right side of the equation is just x^5, it means the entire expression: (1+x(1+x(1+x(1+x)))) should get canceled out which means we will get the same expression in the numerator as well. You don't need to do it. It is logical since otherwise, you will not get the reduced expression x^5. Also, you can see that you will get something like this in the numerator since the powers are increasing. If you want to see it: A2= x( 1 + x + x^2 + x^3 + x^4) = x( 1 + x(1 + x + x^2 + x^3)) = x( 1 + x(1 + x( 1 + x + x^2))) = x( 1 + x(1 + x( 1 + x( 1 + x)))) Similarly A7 = x^6( 1 + x(1 + x( 1 + x( 1 + x)))) So \frac{A7}{{x(1+x(1+x(1+x(1+x))))}} = \frac{x^6( 1 + x(1 + x( 1 + x( 1 + x))))}{x( 1 + x(1 + x( 1 + x( 1 + x))))}= x^5 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  02 Feb 2011, 07:32
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Its not so hard when you realize how can you solve it, but until that, you spent half of your life.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  24 Jul 2012, 20:36
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rajathpanta wrote:
omg.. by the time u read and digest the question its 1 minut e

It certainly takes you a minute or even more to get through the question and digest it but after that, it takes you less than a minute to solve it. This is true for most GMAT questions. If you understand the question well, it takes you very little time to actually solve it. If you don't understand the question well, you could end up spending 20 mins on it.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 03 Feb 2010 Posts: 68 Followers: 1 Kudos [?]: 29 [0], given: 4 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 31 Mar 2010, 12:19 great explanation. Thanks. +1 to you Manager Joined: 05 Mar 2010 Posts: 218 Followers: 1 Kudos [?]: 24 [0], given: 8 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 01 Apr 2010, 10:41 Is it a GMAT level question?????? _________________ Success is my Destiny SVP Joined: 16 Nov 2010 Posts: 1688 Location: United States (IN) Concentration: Strategy, Technology Followers: 31 Kudos [?]: 330 [0], given: 36 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 04 Feb 2011, 05:00 Hi Karishma What is the meaning of this ? "it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den." Regards, Subhash _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Math Forum Moderator Joined: 20 Dec 2010 Posts: 2036 Followers: 133 Kudos [?]: 1058 [0], given: 376 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 04 Feb 2011, 07:26 A_n=x^{(n-1)}(1+x+x^2+x^3+x^4) x(1+x(1+x(1+x(1+x))))=x(1+x+x^2+x^3+x^4) \frac{x^{(n-1)}(1+x+x^2+x^3+x^4)}{x(1+x+x^2+x^3+x^4)}=x^5 x^{(n-1)}=x^6 n-1=6 n=7 Ans: "B" _________________ Intern Joined: 30 Mar 2011 Posts: 48 Schools: Virginia Tech Followers: 0 Kudos [?]: 2 [0], given: 0 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 13 Apr 2011, 12:25 I hope this is a 700+ level question... Manager Joined: 14 Feb 2011 Posts: 69 Followers: 1 Kudos [?]: 1 [0], given: 2 Re: Tough Problem Solving Infinite Sequence Problem [#permalink] 27 Dec 2011, 12:03 aznboi986 wrote: I hope this is a 700+ level question... this is 800+ level question You definitely do not see any question like this one on real exam. GC problems usually too difficult. Intern Joined: 05 Jun 2011 Posts: 18 Location: Dhaka,Bangladesh Schools: HBS,Yale, British Columbia Business School, Wharton Business School Followers: 1 Kudos [?]: 10 [0], given: 0 Sequence problem [#permalink] 18 Apr 2012, 18:36 In the infinite sequence A, the nth term, A(n), is given by x^(n-1) + x^n+ x^(n+1) + x^(n+2) + x^(n +3) where x is a positive integer constant. For what value of n is the ratio of A(n) to x(1+x(1+x (1+x(1+x)))) equal to x^5? (A) 8 (B) 7 (C) 6 (D) 5 (E) 4 Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 25927 Followers: 4004 Kudos [?]: 36804 [0], given: 4533 Re: Sequence problem [#permalink] 18 Apr 2012, 23:06 Expert's post Ashikurrahman wrote: In the infinite sequence A, the nth term, A(n), is given by x^(n-1) + x^n+ x^(n+1) + x^(n+2) + x^(n +3) where x is a positive integer constant. For what value of n is the ratio of A(n) to x(1+x(1+x (1+x(1+x)))) equal to x^5? (A) 8 (B) 7 (C) 6 (D) 5 (E) 4 Posted from my mobile device Merging similar topics. Please ask if anything remains unclear. _________________ Intern Joined: 28 Jan 2012 Posts: 6 Followers: 0 Kudos [?]: 0 [0], given: 2 Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] 05 Jun 2012, 08:42 i) x(1 + x(1 + x(1 + x(1 + x)))) = x + x^2 + x^3 + x^4 + x^5 ii) An = x^(n-2) (x + x^2 + x^3 + x^4 + x^5) Divide ii by i = x^(n-2) We want ratio = x^5. So, n-2 = 5 => n = 7. (B) Senior Manager Status: Prevent and prepare. Not repent and repair!! Joined: 13 Feb 2010 Posts: 278 Location: India Concentration: Technology, General Management GPA: 3.75 WE: Sales (Telecommunications) Followers: 9 Kudos [?]: 29 [0], given: 282 Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] 24 Jul 2012, 18:27 omg.. by the time u read and digest the question its 1 minut e _________________ I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+ Verbal Forum Moderator Joined: 23 Oct 2011 Posts: 286 Followers: 32 Kudos [?]: 453 [0], given: 19 Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] 30 Jul 2012, 21:28 Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice. _________________ ******************** Push +1 kudos button please, if you like my post. Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 77 Kudos [?]: 576 [0], given: 43 Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] 31 Jul 2012, 00:49 mohankumarbd wrote: Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice. Who can tell you? If you ask all those who took the test if they ever saw such a question on a real test, you might get the real picture... IMO, the chance is slim that such a question will appear on a real test. It is too technical, too lengthy to be done with plugging in numbers... Until now, I didn't get the feeling that GMAT wants to test just algebraic abilities. Not that this question needs some really advanced techniques, but it's above basics... _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5255 Location: Pune, India Followers: 1271 Kudos [?]: 6302 [0], given: 173 Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] 31 Jul 2012, 07:36 Expert's post mohankumarbd wrote: Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice. It is an algebra question that looks tricky but can be easily reasoned out. It will take you some time to understand the question but once you do, you can solve it quickly - pretty much like high level GMAT questions. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  31 Jul 2012, 08:35
Great Explanation +1
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X   [#permalink] 31 Jul 2012, 08:35

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