December 17, 2018 December 17, 2018 06:00 PM PST 07:00 PM PST Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong. December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 24 Jan 2017
Posts: 146
GMAT 1: 640 Q50 V25 GMAT 2: 710 Q50 V35
GPA: 3.48

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
06 Jun 2017, 04:35
sonikavadhera wrote: Bunuel wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
Given that: \(a_2=a_1*k\) \(a_3=a_1*k^2\) \(a_4=a_1*k^3\) ... \(a_n=a_1*k^{n1}\)
Also given that k>0 and n=9.
(1) The ninth term in this sequence is 81 > \(a_9=a_1*k^8=81\). If \(a_1=1\), then all but \(a_1\) will be greater than 1, but if \(a_1=2\), then all will be greater than 1. Not sufficient.
(2) The fifth term in this sequence is 1 > \(a_5=a_1*k^4=1\). Now, if \(a_1<1\), then \(k>1\), and all terms from \(a_5\) (\(a_6\), \(a_7\), \(a_8\), and \(a_9\)), so 4 terms will be greater than 1 AND if \(a_1>1\), then \(k<1\), and all terms till \(a_5\) (\(a_2\), \(a_3\) and \(a_4\)), so again 4 terms will be greater than 1. Sufficient.
Answer: B. I am not sure how statement 2 is sufficient if k is a fraction.? Hi, The reasoning of Bunuel about statement 2 is still correct considering k is a fraction. In fact, Bunuel did not eliminate this consideration (that k is a fraction).  If k > 1 (no matter whether k is a fraction or not), then all 4 terms to the right of 5th term will be >1. It also means that all 4 terms to the left hand of 5th term will be <1. Ex: k=3/2, then we have the following sequence: 16/81; 8/27; 4/9; 2/3; 1; 3/2; 9/4; 27/8; 81/16 . You can see that 4 terms to the right of 5th term are 3/2; 9/4; 27/8; 81/16 => All these 4 terms are >1  Similarly, If k < 1 (no matter whether k is a fraction or not), then all 4 terms to the right of 5th term will be <1. It also means that all 4 terms to the left hand of 5th term will be >1. If you still have concern(s), just indicate why you think statement 2 is insufficient if k is fraction, then I will response accordingly.



Intern
Joined: 18 Apr 2013
Posts: 34

The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
23 Jul 2017, 03:08
Bunuel wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
Given that: \(a_2=a_1*k\) \(a_3=a_1*k^2\) \(a_4=a_1*k^3\) ... \(a_n=a_1*k^{n1}\)
Also given that k>0 and n=9.
(1) The ninth term in this sequence is 81 > \(a_9=a_1*k^8=81\). If \(a_1=1\), then all but \(a_1\) will be greater than 1, but if \(a_1=2\), then all will be greater than 1. Not sufficient.
(2) The fifth term in this sequence is 1 > \(a_5=a_1*k^4=1\). Now, if \(a_1<1\), then \(k>1\), and all terms from \(a_5\) (\(a_6\), \(a_7\), \(a_8\), and \(a_9\)), so 4 terms will be greater than 1 AND if \(a_1>1\), then \(k<1\), and all terms till \(a_5\) (\(a_2\), \(a_3\) and \(a_4\)), so again 4 terms will be greater than 1. Sufficient.
Answer: B. Hi Bunuel, I don't quite understand the explanation for statement 2. You mention " if a1>1, then k<1, and all terms till a5(a2, a3 and a4), so again 4 terms will be greater than 1. Sufficient." Why would K be <1? In the question it is clearly stated that K is a positive constant and does not equal 1. Thus, K will always be more than 1. Please correct me if I am wrong. Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
23 Jul 2017, 04:56
roastedchips wrote: Bunuel wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
Given that: \(a_2=a_1*k\) \(a_3=a_1*k^2\) \(a_4=a_1*k^3\) ... \(a_n=a_1*k^{n1}\)
Also given that k>0 and n=9.
(1) The ninth term in this sequence is 81 > \(a_9=a_1*k^8=81\). If \(a_1=1\), then all but \(a_1\) will be greater than 1, but if \(a_1=2\), then all will be greater than 1. Not sufficient.
(2) The fifth term in this sequence is 1 > \(a_5=a_1*k^4=1\). Now, if \(a_1<1\), then \(k>1\), and all terms from \(a_5\) (\(a_6\), \(a_7\), \(a_8\), and \(a_9\)), so 4 terms will be greater than 1 AND if \(a_1>1\), then \(k<1\), and all terms till \(a_5\) (\(a_2\), \(a_3\) and \(a_4\)), so again 4 terms will be greater than 1. Sufficient.
Answer: B. Hi Bunuel, I don't quite understand the explanation for statement 2. You mention " if a1>1, then k<1, and all terms till a5(a2, a3 and a4), so again 4 terms will be greater than 1. Sufficient." Why would K be <1? In the question it is clearly stated that K is a positive constant and does not equal 1. Thus, K will always be more than 1. Please correct me if I am wrong. Thanks! We are told that k is positive constant and k ≠ 1. So, 0 < k < 1 or k > 1. Nowhere we are told that k is an integer.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 649

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
23 Jul 2017, 14:48
akijuneja wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1? (1) The ninth term in this sequence is 81. (2) The fifth term in this sequence is 1 I'm going to try to explain this one in a less 'mathematical' way, as an alternative for those of us who aren't super enthusiastic about algebra. Sequences are also pretty easy to translate into plain English, and that's how I prefer to think about them. The question stem says that we're going to build a sequence by starting at some number, then multiplying by the same value over and over until we have 9 terms. For example, this would fit the rules: 1, 3, 9, 27, 81, 243, ... etc. This would also fit the rules: 4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/32, ... etc. This would not fit the rules, since k can't equal 1 (we can't just multiply by 1 over and over): 3, 3, 3, 3, 3, 3, ...This also wouldn't work, since we have to multiply by the same value instead of adding the same value: 1, 2, 3, 4, 5, ...So that's what the question is telling us. Notice that it's not telling us what the first number in the sequence is, so that could be literally anything. The first number could even be negative! (We just can't multiply by a negative number to get the next term.) Finally, the question itself asks how many of the first 9 terms are greater than 1. I'm not sure how to figure that out mathematically, to be honest, so I'm going to focus on case testing. Now, the statements. Statement 1 The ninth term is 81. _ _ _ _ _ _ _ _ 81 I know that I started with some value (which I don't know) and then multiplied by the same number over and over, until I hit 81 after multiplying 8 times. In order to go backwards, I'll have to start with the 81, and divide by the same number, over and over. Suppose that k = 3. Build the sequence backwards: _ _ _ _ _ _ _ 27 81 _ _ _ _ _ _ 9 27 81 _ _ _ _ _ 3 9 27 81 _ _ _ _ 1 3 9 27 81 The other four terms will be less than 1. So the answer to the question is "4 terms". Now let's try a really crazy case. Suppose that k = 81. That is, we'll be dividing by 81 over and over. _ _ _ _ _ _ _ 1 81 _ _ _ _ _ _ 1/81 1 81 1/81 is less than 1, and so are the other 6 terms at the beginning. The answer the question is "7 terms". This statement is not sufficient. Statement 2: The fifth term of the sequence is 1. So, it looks like this: _ _ _ _ 1 _ _ _ _ We'll be going in both directions now. To go to the left, we'll divide by k. To go to the right, we'll multiply by k. Let's say that k = 2. _ _ 1/4 1/2 1 2 4 _ _ Interesting. The four to the left are less than 1. The answer to the question is "4". Let's say that k = 1/2. _ _ 4 2 1 1/2 1/4 _ _ Okay, now the four to the right are less than 1. The answer to the question is "4". Will that always be true? Yeah  if k is a fraction, then the four numbers to the right will all be smaller than 1. If k isn't a fraction, the four numbers to the left will all be smaller than 1. That's because we're either dividing 1 by a number repeatedly (making it smaller), or multiplying 1 by a number repeatedly (making it bigger). That's why this statement is sufficient, even if k is a fraction.
_________________



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4305
Location: United States (CA)

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
08 Dec 2017, 11:59
akijuneja wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1? (1) The ninth term in this sequence is 81. (2) The fifth term in this sequence is 1 We are given that the next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1, and we need to determine how many of the first nine terms in this sequence are greater than 1. Statement One Alone: The ninth term in this sequence is 81. Without knowing any other terms in the sequence or the value of the positive constant k we are multiplying, we can’t determine the number of terms in the sequence that are greater than 1. For example, if k = 3, then by going backward we have: 9th term = 81, 8th term = 27, 7th term = 9, 6th term = 3, 5th term = 1, 4th term = 1/3, and so on. In this case, we have 4 terms that are greater than 1. However, if k = 9, then by going backward again we have: 9th term = 81, 8th term = 9, 7th term = 1, 6th term = 1/9, and so on. In this case we have only 2 terms that are greater than 1. Therefore, statement one alone is not sufficient. We can eliminate answer choices A and D. Statement Two Alone: The fifth term in this sequence is 1. Since there are 9 total terms, we see that the 5th term is the middle term in our sequence. In other words there are 4 terms below the 5th term and 4 terms above the 5th term. Since we know that k is positive, it’s either a positive proper fraction or a number greater than 1. Thus, regardless of whether k is a positive proper fraction or a number greater than 1, there will be 4 numbers above 1 and 4 numbers below 1. For instance, if k = 1/2, then the 1st to the 4th terms, inclusive, are greater than 1, and if k = 2, then the 6th to the 9th terms, inclusive, are greater than 1. Either way, there are 4 terms greater than 1. Statement two alone is sufficient to answer the question. Answer: B
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 19 Aug 2016
Posts: 84

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
22 Apr 2018, 18:03
Bunuel wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
Given that: \(a_2=a_1*k\) \(a_3=a_1*k^2\) \(a_4=a_1*k^3\) ... \(a_n=a_1*k^{n1}\)
Also given that k>0 and n=9.
(1) The ninth term in this sequence is 81 > \(a_9=a_1*k^8=81\). If \(a_1=1\), then all but \(a_1\) will be greater than 1, but if \(a_1=2\), then all will be greater than 1. Not sufficient.
(2) The fifth term in this sequence is 1 > \(a_5=a_1*k^4=1\). Now, if \(a_1<1\), then \(k>1\), and all terms from \(a_5\) (\(a_6\), \(a_7\), \(a_8\), and \(a_9\)), so 4 terms will be greater than 1 AND if \(a_1>1\), then \(k<1\), and all terms till \(a_5\) (\(a_2\), \(a_3\) and \(a_4\)), so again 4 terms will be greater than 1. Sufficient.
Answer: B. Hi Bunuel, Why are we even considering k<1 as the question explicitly says pos constant...? Also can u please give me an example where k>1 and a>1 gives a^5=1



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
22 Apr 2018, 20:31
zanaik89 wrote: Bunuel wrote: The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?
Given that: \(a_2=a_1*k\) \(a_3=a_1*k^2\) \(a_4=a_1*k^3\) ... \(a_n=a_1*k^{n1}\)
Also given that k>0 and n=9.
(1) The ninth term in this sequence is 81 > \(a_9=a_1*k^8=81\). If \(a_1=1\), then all but \(a_1\) will be greater than 1, but if \(a_1=2\), then all will be greater than 1. Not sufficient.
(2) The fifth term in this sequence is 1 > \(a_5=a_1*k^4=1\). Now, if \(a_1<1\), then \(k>1\), and all terms from \(a_5\) (\(a_6\), \(a_7\), \(a_8\), and \(a_9\)), so 4 terms will be greater than 1 AND if \(a_1>1\), then \(k<1\), and all terms till \(a_5\) (\(a_2\), \(a_3\) and \(a_4\)), so again 4 terms will be greater than 1. Sufficient.
Answer: B. Hi Bunuel, Why are we even considering k<1 as the question explicitly says pos constant...? Also can u please give me an example where k>1 and a>1 gives a^5=1 1. We are told that k is positive constant and k ≠ 1. Nowhere we are told that k is an integer. So, 0 < k < 1 or k > 1. 2. I guess, you are talking about the second statement. \(a_5=a_1*k^4=1\) cannot be true if both k and a1 are greater than 1. That's why the solution considers two cases: \(a_1<1\), \(k>1\) and \(a_1>1\), \(k<1\).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 02 Oct 2017
Posts: 728

Re: The next number in a certain sequence is defined by multiply
[#permalink]
Show Tags
09 May 2018, 03:52
K and a are positive as a is positive I) 9th term ak^8=81 Here we don't know which term is 1 so we can't say how many terms are greater than 1 Eg a=1/(81)^7 k=81 only one term greater than 1 a=1/(9)^6 k=9 two terms greater than 1 So there can be multiple cases so insufficient II) 5th term ak^4=1 K is positive so every term after fifth would be greater than 1 So 4 terms are greater than 1 Sufficient Posted from my mobile device
_________________
Give kudos if you like the post




Re: The next number in a certain sequence is defined by multiply &nbs
[#permalink]
09 May 2018, 03:52



Go to page
Previous
1 2
[ 28 posts ]



