Last visit was: 21 May 2024, 19:10 It is currently 21 May 2024, 19:10
Close
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
Your Progress

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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Intern
Intern
Joined: 16 Aug 2015
Posts: 7
Own Kudos [?]: 8 [5]
Given Kudos: 107
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625654 [1]
Given Kudos: 81918
Send PM
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6817
Own Kudos [?]: 30292 [1]
Given Kudos: 799
Location: Canada
Send PM
Manager
Manager
Joined: 10 Apr 2018
Posts: 186
Own Kudos [?]: 450 [0]
Given Kudos: 115
Location: United States (NC)
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Hi

Well we all understood that series is \(\frac{1}{2}\) , \(\frac{-1}{2^2}\) , \(\frac{1}{2^3}\), \(\frac{-1}{2^4}\), \(\frac{1}{2^5}\) ,\(\frac{-1}{2^6}\),\(\frac{1}{2^7}\) ,\(\frac{-1}{2^8}\) ,\(\frac{1}{2^9}\) ,\(\frac{-1}{2^10}\)

Can i rewrite the series as

{\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } - { \(\frac{1}{2^2}\) + \(\frac{1}{2^4}\) + \(\frac{1}{2^6}\)+ \(\frac{1}{2^8}\) + \(\frac{1}{2^10}\) }

{\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } -\(\frac{1}{2}\) { \(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) }

Let {\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } =q
then
\(q-\frac{q}{2}\)

\(\frac{q}{2}\)

now q is in GP and we need to find the sum of GP whose first term is \(\frac{1}{2}\) and r is \(\frac{1}{4}\) , n is 5
Either you can use the formula for sum of infinte terms of GP or genereal formula .

Lets try the general formula which says sum of n terms of GP

we have \(\frac{\frac{1}{2} * (1 - \frac{1}{4^5})}{ (1-\frac{1}{4})}\)

\(\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}\)

Can we say \(\frac{4^5-1}{4^5}\) is nearly equal to 1

So this expression \(\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}\) hence simplifies to \(\frac{2}{3}\)

now q= \(\frac{2}{3}\)

we need the value of \(\frac{q}{2}\) = \(\frac{1}{3}\)

So Choice D suits our Answer.

PS: Does this solution look long - Yeah because i tried to explain each step. I am sure we will not being many steps that i wrote down . I did get my solution in 1: 03 min


abhimahna

I guess you missed on signs in this post , lucky you ended getting the right answer.

https://gmatclub.com/forum/kth-term-of- ... l#p1724824
CEO
CEO
Joined: 07 Mar 2019
Posts: 2576
Own Kudos [?]: 1835 [1]
Given Kudos: 763
Location: India
WE:Sales (Energy and Utilities)
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
1
Bookmarks
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


Series has 10 terms where first term \(= \frac{1}{2}\) and 10th term \(= - \frac{1}{2^{10}}\).
Every odd is positive and every even term is negative.

The series becomes
\(\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}\)

Separating positive terms and negative terms we have
\(\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}\)
\(= \frac{1}{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) - \frac{1}{4}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)\)
\(= (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) * (\frac{1}{2} - \frac{1}{4})\)
\(= \frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)\)

And
\(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} .. > 1.25\)
\(\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > \frac{1}{4} * 1.25\)
\(\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > 0.31\)

Checking answers only D is correct since \(\frac{1}{4} < 0.31 < \frac{1}{2}\)

Answer (D).
Senior Manager
Senior Manager
Joined: 27 Mar 2017
Posts: 272
Own Kudos [?]: 77 [0]
Given Kudos: 406
Location: Saudi Arabia
GMAT 1: 700 Q47 V39
GPA: 3.36
Send PM
For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Bunuel wrote:

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 \(sum=\frac{b}{1-r}\), 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.


Bunuel How do we know that the sum of this sequence will be very close to 1/3 ?
GMAT Club Legend
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21843
Own Kudos [?]: 11690 [0]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Expert Reply
altairahmad wrote:
Bunuel wrote:

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 \(sum=\frac{b}{1-r}\), 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.


Bunuel How do we know that the sum of this sequence will be very close to 1/3 ?


Hi altairahmad,

Bunuel's explanation included a formula that shows why the sum gets fairly close to 1/3. Looking at the answer choices though, you do NOT actually have to get to that conclusion (since the answers are RANGES, we just need to do enough work to define which "range" the total will be in - and then we can stop working).

By plugging in the first few numbers (1, 2, 3, 4) into the sequence, you can see that a pattern emerges among the terms....

1st term = +1/2
2nd term = -1/4
3rd term = +1/8
4th term = -1/16

The terms follow a positive-negative-positive-negative pattern all the way to the 10th term and each term is the product of the prior term and 1/2. By "pairing up' the terms, another pattern emerges....

1/2 - 1/4 = 1/4

1/8 - 1/16 = 1/16

1/32 - 1/64 = 1/64

Etc.

The "pairs" get progressively smaller (notice how each is the product of the prior term and 1/4). This means that we're "starting with" 1/4 and adding progressively TINIER fractions to it. Since we're just adding 4 fractions (that are getting REALLY SMALL) to 1/4, this means that we're going to end up with a total that's just a LITTLE MORE than 1/4. Looking at the answer choices, there's only one answer that fits:

GMAT assassins aren't born, they're made,
Rich
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5351
Own Kudos [?]: 4013 [0]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
prathns 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 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


Given: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((-1)^{(k+1)}*(\frac{1}{2^k})\).
Asked: If T is the sum of the first 10 terms in the sequence, then T is

\(T = \frac{1}{2} - \frac{1}{4} + \frac{1}{8 }- \frac{1}{16} + ......- \frac{1}{2^{10}}\)

\(T_{max} = \frac{1}{2} - \frac{1}{4} + ...... to\ infinity = \frac {\frac{1}{2} }{1+ \frac{1}{2}} =\frac{1}{3} \)

IMO D
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18886
Own Kudos [?]: 22291 [4]
Given Kudos: 285
Location: United States (CA)
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
3
Kudos
1
Bookmarks
Expert Reply
prathns 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 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


Solution:
We are given that for every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as exact values. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let’s start by listing the first four terms.
k = 1:
(-1)^(1+1) x (1/2^1)
(-1)^2 x 1/2
1 x 1/2 = 1/2
k = 2:
(-1)^(2+1) x (1/2^2)
(-1)^3 x 1/4
-1 x 1/4 = -1/4
k = 3:
(-1)^(3+1) x (1/2^3)
(-1)^4 x 1/8
1 x 1/8 = 1/8
k = 4:
(-1)^(4+1) x (1/2^4)
(-1)^5 x 1/16
-1 x 1/16 = -1/16
Recall that we are trying to estimate the value of T = 1/2 + (-1/4) + 1/8 + (-1/16) + … until we have 10 terms. In other words, T = 1/2 – 1/4 + 1/8 – 1/16 + … until there are 10 terms.
We should notice that the absolute values of the terms are getting smaller:
|1/2|>|-1/4|>|1/8|>|-1/16|.
Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2) but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because ½ and -1/4 are the largest term and the smallest term, respectively, in our set, the sum will never fall below ¼ or exceed ½.
Thus, we conclude that T is greater than 1/4 but less than 1/2.

Answer: D
Tutor
Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Posts: 452
Own Kudos [?]: 776 [0]
Given Kudos: 8
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Expert Reply
prathns 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 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



Let us first look at the terms by substituting k = 1, 2, 3, 4 ... :
1/2, -1/4, 1/8, -1/16 ...

Observe that the options are in ranges and so, we need not necessarily calculate the exact value of this summation. Also, both 1/2 and 1/4 are there in the options - the first 2 terms of the series above

Adding the first 2 terms, we get 1/2 - 1/4 = 1/4
Thereafter, the sum of every pair is a positive quantity, for example, 1/8 - 1/16 > 0 etc.
Thus, the sum is greater than 1/4

Again, leaving out the first tern, 1/2, if we pair, each pair would be negative (for example: -1/4 + 1/8 < 0) and the 10th term would be left over, which also is negative.
Thus, the sum is less than 1/2

Answer D
Tutor
Joined: 02 Oct 2015
Posts: 161
Own Kudos [?]: 23 [0]
Given Kudos: 4
Send PM
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Expert Reply
­Track the back and forth along a number line and it's easy to see where it'll land:

GMAT Club Bot
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
   1   2   3 
Moderator:
Math Expert
93373 posts