Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 01 Aug 2009
Posts: 19

For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
07 Jan 2010, 05:22
21
This post received KUDOS
103
This post was BOOKMARKED
Question Stats:
61% (01:19) correct 39% (01:36) wrong based on 2014 sessions
HideShow timer Statistics
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence then T is A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
My GMAT JOURNEY : 500 to 640 in a Month!!  > http://gmatclub.com/forum/finallydonegmat1500togmat2640inamonth89600.html#p678289



Math Expert
Joined: 02 Sep 2009
Posts: 43862

For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
07 Jan 2010, 06:30
55
This post received KUDOS
Expert's post
111
This post was BOOKMARKED
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T isA. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4 First of all we see that there is set of 10 numbers and every even term is negative. Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now. And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. BUT there is shortcut:Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\). Now, the sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term.So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\) This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out). So the answer is D. Answer: D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661
_________________
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



Manager
Joined: 27 Apr 2008
Posts: 191

Re: Sequence [#permalink]
Show Tags
07 Jan 2010, 08:13
22
This post received KUDOS
17
This post was BOOKMARKED
Alternative, if you use the geometric series formula.
S = \(\frac{a(1r^n)}{1r}\)
where a = first term, r = multiple factor, n = # of terms.
So in this case, a = 1/2, r = 1/2, n = 10.
Therefore,
S=( (1/2) * (1 (1/2)^10) ) / (1(1/2) S=( (1/2) * (1  (1/1024) ) / (3/2) S=(11/1024) / 3 S=(1023/1024 ) / 3 Since 1023/1024 is close to 1, dividing it by 3 would get us to approximately 1/3, which is between 1/2 and 1/4. So the answer is D.



Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 290
Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)

Re: Sequence [#permalink]
Show Tags
27 Apr 2010, 18:08
Nice question and nice explanations! I was more leaning toward geometric series solution.
_________________
"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??



Manager
Joined: 03 May 2010
Posts: 86
WE 1: 2 yrs  Oilfield Service

Re: DS : Number Theory [#permalink]
Show Tags
10 Jul 2010, 00:10
8
This post received KUDOS
2
This post was BOOKMARKED
Kth term: A(k) = (1) ^ (k+1) * 1 / 2^k
T is the sum of the first 10 terms of this sequence.
A(1) = (1)^(1+1)*1/2^1 = 1/2
A(2) = (1)^(1+2)*1/2^2 = 1/4
You can see that for odd k it would be positive values of \(1/2^k\) and for even k it would be negative values of \(1/2^k\)
So it's like this:
\(T = \frac{1}{2}  \frac{1}{4} + \frac{1}{18}  \frac{1}{16}+ ...\)
Now look for a pattern in the sums of various number of terms.
i.e. Sum of first 2 terms = 1/2  1/4 = 1/4 Sum of first 3 terms = 1/4 + 1/8 = 3/8 Sum of first 4 terms = 3/8  1/16 = 5/16 Sum of first 5 terms = 5/16 + 1/32 = 11/32 and so on...
You can see that all these sums are between 1/4 and 1/2.
On the GMAT when you have tested such a question for about half the terms and you have established a pattern you can safely assume it will continue. This is because this series is analogous to how the GMAT itself adapts the difficulty of its questions based on your answers. It sort of zigzags in a diminishing pattern. So as you increase k, you will be varying only slightly around a sort of ultimate stagnant value at infinity which you could determine as \(\frac{a}{(1  r)}\) i.e. the sum of an infinite Geometric Progression with first term a and common ratio r.
Here we have first term 1/2 and common ratio 1/2
=> Infinite sum = \(\frac{1}{2*(1  (1/2))}= \frac{1}{2*1.5} = \frac{1}{3}\)
With a large number of terms such as 10 terms, your series sum would actual TEND to stagnate around this infinite sum.
Pick D



Manager
Joined: 30 May 2010
Posts: 189

Re: Sequence problem GMATPrep first 10 terms [#permalink]
Show Tags
17 Aug 2010, 23:17
19
This post received KUDOS
11
This post was BOOKMARKED
Initially, I thought about calculating every term. This is almost always the wrong approach. It is more of a number properties problem than sequences.
For the first term, it alternates between positive and negative. For even k, it is positive 1 * \(1/2^k\) and negative 1 for odd k.
The first term is \(1 * 1/2 = 1/2\)
The second term is \(1 * 1/4 = 1/4\)
\(1/2  1/4 = 1/4\)
The third term is \(1 * 1/8 = 1/8\)
\(1/4 + 1/8 = 3/8\)
Looking at the answer choices, you don't need to continue. Since the denominator is increasing exponentially, the terms added and subtracted are becoming closer to 0. From the first term, we know we will never go above \(1/2\). After subtracting the second term, we know we will never go below \(1/4\).



Manager
Joined: 06 Aug 2010
Posts: 217
Location: Boston

Re: Sequence [#permalink]
Show Tags
01 Nov 2010, 10:41
6
This post received KUDOS
1
This post was BOOKMARKED
prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is
a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.
I have no clue what info has been given and how to use it to derive T.
Kindly post a detailed explanation.
Thanks. Prath. When doing sequence problems, it usually helps to look at at least the first few terms. So in this case: \(a_k = (1)^{k+1} * \frac{1}{2^k}\) This gives us: \(a_1 = \frac{1}{2}\) \(a_2 = \frac{1}{4}\) \(a_3 = \frac{1}{8}\) \(a_4 = \frac{1}{16}\) We can stop there. The first, and largest, term is 1/2, and we then subtract 1/4. We will then add and subtract fractions that will continue to get smaller and smaller. So we can immediately eliminate A, B, and C, since the sum cannot possibly be greater than 1/2. Now we're left with D and E. Note that the sum of the first two terms is 1/4. Then you add another 1/8 and subtract 1/16. This pattern will continue all the way through the tenth term, and you should be able to see that there's no way this sum will become less than 1/4. So the answer is D.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7951
Location: Pune, India

Re: Sequence [#permalink]
Show Tags
26 Feb 2011, 20:46
2
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is
a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.
I have no clue what info has been given and how to use it to derive T.
Kindly post a detailed explanation.
Thanks. Prath. Using some keen observation, you can quickly arrive at the answer... Terms will be: \(\frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32}  ...  \frac{1}{1024}\) For every pair of values: \(\frac{1}{2}  \frac{1}{4} = \frac{1}{4}\) \(\frac{1}{8}  \frac{1}{16} = \frac{1}{16}\) etc... So this series is actually just \(\frac{1}{4} + \frac{1}{16} + ... + \frac{1}{1024}\) So the sum is definitely greater than 1/4. When you add an infinite GP with 1/16 as first term and 1/4 as common ratio, the sum will be \(\frac{\frac{1}{16}}{1\frac{1}{4}} = 1/12\). Here, the sum of terms 1/16 + 1/64 + ... 1/1024 is definitely less than 1/12. Hence the sum is definitely less than 1/2. Answer is (D).
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Status: pursuing a dream
Joined: 02 Jun 2011
Posts: 43
Schools: MIT Sloan (LGO)

Re: Sequence [#permalink]
Show Tags
13 Aug 2011, 11:42
24
This post received KUDOS
2
This post was BOOKMARKED
Hello! I was just doing this problem and though I would add some "graphic" approach in case it is useful for someone as it has been for me. You can easily draw a graph of the first points on any series to see if they follow a regular pattern. This one specifically is extremely easy to catch as you draw 3+ points, something like this:
Attachments
GP.jpg [ 18.77 KiB  Viewed 67102 times ]



Manager
Status: On...
Joined: 16 Jan 2011
Posts: 181

Re: Sequence [#permalink]
Show Tags
19 Aug 2011, 15:31
4
This post received KUDOS
I liked Graphical approach is a good one. BUT  felt there is no need of any calculations T = +(1/2 + 1/2^3 + 1/2^5 + 1/2^7 +1/2^9) (1/2^2 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10)Subtract the top and bottom T = 1/4 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10Now I know T is greater than 1/4 , but if sum of the remaining is smaller than 1/4, then it will definitely be between 1/4 and 1/2 T = 1/4 + 1/4(1/2^2 + 1/2^4 + 1/2^6 + 1/2^8)So sum of the remaining is smaller than 1/4, so ans. is between 1/4 and 1/2 OA D
_________________
Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudoswhataretheyandwhywehavethem94812.html



Senior Manager
Joined: 30 Aug 2009
Posts: 280
Location: India
Concentration: General Management

Re: Integer K from 1 to 10 [#permalink]
Show Tags
12 Jan 2012, 03:35
1
This post received KUDOS
enigma123 wrote: For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)] × (1 / 2^k). If T is the sum of the first 10 terms of the sequence, then T is: A. greater than 2 B. between 1 and 2 C. between 1/2 and 1 D. between 1/4 and ½ E. less than 1/4
Can someone please let me know what is the concept behind this and how to solve? Every odd value of k will make (k+1) even and [(1)^(k+1)] will be +ve so we will have k = 1, [(1)^(1+1)] × (1 / 2^1) = 1*(1/2) = 1/2 k = 2, [(1)^(2+1)] × (1 / 2^2) = 1*1/4 = 1/4 k = 3, [(1)^(3+1)] × (1 / 2^3) = 1 * 1/8 = 1/8 so we can observe we will get alternate +ve and ve value for (1/2)^k (1/2)  (1/4) + (1/8)  (1/16) + (1/32)  (1/64) + (1/128)  (1/256) + (1/512)  (1/1024) [(1/2)+ (1/8) + (1/32)+ (1/128) + (1/512)]  [(1/4)+(1/16)+(1/64)+(1/256)+(1/1024)] = approx 0.3 so D



Manager
Joined: 16 Feb 2012
Posts: 213
Concentration: Finance, Economics

Re: Sequence [#permalink]
Show Tags
26 Apr 2012, 06:55
Bunuel wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T isA. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4 First of all we see that there is set of 10 numbers and every even term is negative. Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now. And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. BUT there is shortcut:Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\). Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\) This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out). So the answer is D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)?
_________________
Kudos if you like the post!
Failing to plan is planning to fail.



Math Expert
Joined: 02 Sep 2009
Posts: 43862

Re: Sequence [#permalink]
Show Tags
26 Apr 2012, 07:07



Intern
Joined: 17 Jan 2012
Posts: 12

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
27 Oct 2012, 06:44
1
This post received KUDOS
simple explanation: look at the first term and second term: it says quarter is removed from half, result would be less than half of first term.look at the next two terms: already the terms are less than half and quarter, result would much more less than quarter ( half of the first term in the sequence terms). similarity, adding all the results, no mater how many terms are there, ultimate result would be always less than 1/2 and more than 1/4. but, if there are more terms it may happen that result would much more more than small term. example more than small term 1/16.



VP
Joined: 08 Jun 2010
Posts: 1377

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
17 Nov 2012, 02:05
gmat dose not test the memory of formular. HOw to solve in 2 minutes. ?
_________________
visit my facebook to help me. on facebook, my name is: thang thang thang



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7951
Location: Pune, India

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
18 Nov 2012, 10:15
thangvietnam wrote: gmat dose not test the memory of formular.
HOw to solve in 2 minutes. ? Check out the explanation of this question here: http://www.veritasprep.com/blog/2012/03 ... sequences/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Joined: 29 Nov 2012
Posts: 854

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
23 Dec 2012, 23:32
some really good explanations for this question! I was wondering do we need to know the infinite sequence formula  you can solve the question in a minute if you know the formula. The graphic approach was really good too...
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



Math Expert
Joined: 02 Sep 2009
Posts: 43862

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
Show Tags
23 Dec 2012, 23:39



Manager
Joined: 26 Feb 2013
Posts: 53
Concentration: Strategy, General Management
WE: Consulting (Telecommunications)

Re: Numbers , Squences , Indices [#permalink]
Show Tags
22 Apr 2013, 23:47
First term in the series is 1/2, second 1/4 third 1/8 fourth is 1/16. and likewise till 10th term.
so sum of the series is = 1/21/4+1/81/16....
so it is essentially the sum of differences between two adjcent terms((1/21/4)+(1/81/16) till 10th term.
so it will be 1/4 + 1/16 + etc.. we can stop calculating after this. 1/4 = .25 and 1/16 = .0625 and all the subsequent terms will be much less than .0625.
so the sum will be .25+.0625+etc.. = .3abcd
hence its between 1/4 and 1/2
it took me around a minute to solve it. its a sub 1.5 minute question i believe.



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1121
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: Numbers , Squences , Indices [#permalink]
Show Tags
22 Apr 2013, 23:51
2
This post received KUDOS
kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
a. Greater than 2 b. Between 1 and 2 c. Between 1/2 and 1 d.Between 1/4 and 1/2 e.Less than 1/4
I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ? Here we need to find a pattern \(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \(\frac{1}{4}\), the second pair is \(\frac{1}{8}\frac{1}{16}\) positive so we add value to \(\frac{1}{4}\), so the sum will be greater.(this is true also for the next pairs, so we add to \(\frac{1}{4}\) a positive value for each pair) D Hope its clear, let me know
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]




Re: Numbers , Squences , Indices
[#permalink]
22 Apr 2013, 23:51



Go to page
1 2 3
Next
[ 57 posts ]



