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

 It is currently 18 Nov 2019, 20:31

### 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

# For every integer k from 1 to 10, inclusive the "kth term of a certain

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

### Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9788
Location: Pune, India
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

### Show Tags

23 Feb 2017, 01:16
cuhmoon wrote:
Hi Bunuel,

Can we say that if |r| <1, then it's an infinite GP ? How does one define infinite GP?
When does one use the forumula

Sum = b1 (r^n-1)/r-1 ?

Bunuel wrote:
Stiv wrote:
$$\frac{first \ term}{1-constant}$$ Is this formula reversed when we have an increase by 0<constant<1? Does it look like this $$\frac{first \ term}{1+constant}$$?

The sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

To add to what Bunuel said, r is the common ratio and has nothing to do with whether a progression is infinite or finite. In either case, r can be less than 1 or more than 1

e.g.
1, 3, 9, 27, 81 ... (infinite sequence with r = 3)
27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
1, 3, 9 (finite sequence with 3 elements and r = 3)
27, 9, 3 (finite sequence with 3 elements and r = 1/3)

Now the point is that the sum of all terms of the first sequence is infinite. The terms will can getting larger and will keep adding up. So the sum will be infinite.

We can find the exact sum of the rest of the 3 sequences.

27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
Sum = a/(1 - r) = 27/(1 - 1/3) = 81/2

1, 3, 9 (finite sequence with 3 elements and r = 3)
Sum = a(r^n - 1)/(r - 1) = 1*(3^3 - 1)/(3 - 1) = 13

27, 9, 3 (finite sequence with 3 elements and r = 1/3)
Sum = a(1 - r^n)/(1 - r) = 27*(1 - 1/3^3)/(1 - 1/3) = 26*3/2 = 39
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8412
Location: United States (CA)
Re: For every integer k from 1 to 10, inclusive the  [#permalink]

### Show Tags

14 Jun 2017, 16:18
bekbek 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

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 an exact value. 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 out 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 1/2 and -1/4 are the largest term and the smallest term in our set, respectively, the sum will never fall below 1/4 or exceed 1/2.

Thus, we conclude that T is greater than 1/4 but less than 1/2.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Joined: 16 Aug 2015
Posts: 8
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

### Show Tags

15 Jun 2017, 22:26
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 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

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

Other solutions at: http://gmatclub.com/forum/sequence-can- ... ml#p668661

Bunuel, how come the ratio here is -1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks
Retired Moderator
Joined: 17 Jun 2016
Posts: 499
Location: India
GMAT 1: 720 Q49 V39
GMAT 2: 710 Q50 V37
GPA: 3.65
WE: Engineering (Energy and Utilities)
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

### Show Tags

25 Jun 2017, 11:06
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

Refer to the solution in the picture
Attachments

Solution Kth Term.jpeg [ 24.64 KiB | Viewed 822 times ]

_________________
Senior Manager
Joined: 10 Apr 2018
Posts: 267
Location: United States (NC)
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

### Show Tags

26 Apr 2019, 00:54
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
_________________
Probus

~You Just Can't beat the person who never gives up~ Babe Ruth
Senior Manager
Joined: 07 Mar 2019
Posts: 384
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

### Show Tags

10 Sep 2019, 20:19
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 $$\frac{1}{4} < 0.31 < \frac{1}{2}$$

_________________
Ephemeral Epiphany..!

GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
GMATPREPSoft1 680(Q48,V35) June 26, 2019
Non-Human User
Joined: 09 Sep 2013
Posts: 13604
Re: For every integer k from 1 to 10, inclusive, the kth term of  [#permalink]

### Show Tags

07 Oct 2019, 00:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 07 Oct 2019, 00:58

Go to page   Previous    1   2   3   [ 47 posts ]

Display posts from previous: Sort by

# For every integer k from 1 to 10, inclusive the "kth term of a certain

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

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