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

 It is currently 17 Jan 2019, 01:04

### 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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winning strategy for a high GRE score

January 17, 2019

January 17, 2019

08:00 AM PST

09:00 AM PST

Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
• ### Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# For every integer n from 1 to 200, inclusive, the nth term of a certai

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52182
For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

### Show Tags

02 Apr 2018, 01:28
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:13) correct 38% (02:14) wrong based on 70 sessions

### HideShow timer Statistics

For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$. If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

_________________
Retired Moderator
Joined: 25 Feb 2013
Posts: 1220
Location: India
GPA: 3.82
For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

### Show Tags

02 Apr 2018, 08:44
1
1
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$. If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

the sequence can be written as $$\frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3}, \frac{1}{2^4}......................., \frac{-1}{2^{199}}, \frac{1}{2^{200}}$$. For simplicity let $$\frac{1}{2}=a$$

so $$N= (a^2+a^4+.....a^{200})-(a+a^3+......a^{199})$$. This is a GP series. Sum of GP series is $$a_1*\frac{(1-r^n)}{1-r}$$. Here we have two GP series and each has 100 terms, for the first series $$a_1=a^2$$ and for the second series $$a_1=a$$, common ratio, $$r=a^2$$ and $$n=100$$, which is same for both the series.

Hence $$N= a^2*\frac{[1-(a^2)^{100}]}{1-a^2} - a*\frac{[1-(a^2)^{100}]}{1-a^2} = \frac{[1-(a^2)^{100}]}{1-a^2}*(a^2-a)$$. Now substitute the value of $$a$$ here to get

$$[(1-\frac{1}{2^{200}})/(1-\frac{1}{2^2})]*(\frac{1}{2^2}-\frac{1}{2})$$. on solving this equation, we will get $$(\frac{1}{2^{200}}-1)*\frac{1}{3}$$

Among the options given, this value will fall between $$\frac{-1}{2}$$ and $$0$$

Option C
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2722
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

### Show Tags

02 Apr 2018, 23:20
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$. If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

SUGGESTION: For any question of sequence, make sure that you write atleast 4-5 terms to understand the pattern in which the sequence is taking shape.

$$T_1 = -1/2$$
$$T_2 = +1/4$$
$$T_3 = -1/8$$
$$T_4 = +1/16$$
$$T_5 = -1/32$$ and so on...

i.e. Sum = -1/2+1/4-1/8+1/16-1/32+1/64---- and so on

i.e. Sum = -1/4-1/16-1/64 = -0.25-0.0625-0.015625---

i.e. Sum must be between 0 and -1/2

P.S. The formula for sum of a geometric progressions is completely un-necessary for any GMAT question which can be approached using that formula hence reader who haven't seen the formula of sum of a GP need not worry as there is always an alternative that GMAT expects you to know.
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Senior SC Moderator
Joined: 22 May 2016
Posts: 2333
For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

### Show Tags

03 Apr 2018, 08:59
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$. If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

As GMATinsight suggests, because the geometric approach seemed too time-consuming, I looked for a pattern. This arithmetic is not hard.

The nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$

First 6 terms
$$A_1 =(-1^1*2^{-1})=(-1*\frac{1}{2^1})=-\frac{1}{2}$$

$$A_2 =(-1^2*2^{-2})=(1*\frac{1}{2^2})=(1*\frac{1}{4})=\frac{1}{4}$$

$$A_3 =(-1^3*2^{-3})=-\frac{1}{8}$$

$$A_4 =(1*2^{-4})=\frac{1}{16}$$

$$A_5 = (-1*\frac{1}{32})=-\frac{1}{32}$$

$$A_6 = (1*\frac{1}{64})=\frac{1}{64}$$

Pattern? Sum a few terms

We need a pattern for the sum of terms.
Extrapolate directly from the terms' values above, or add:

$$A_1+A_2= (-\frac{1}{2}+\frac{1}{4})=-\frac{1}{4}$$

$$(A_1+A_2) + A_3= (-\frac{1}{4}+\frac{1}{8})=-\frac{1}{8}$$

$$(A_1+A_2+A_3)+ A_4= (-\frac{1}{8}+\frac{1}{16})=-\frac{1}{16}$$

The pattern emerges: Summing the terms gets us closer to 0 by exactly half.

We will never reach 0. We add a positive number to a negative number, but |negative| > positive

$$-Sum + \frac{1}{2}Sum= -\frac{1}{2}Sum$$

The added positive term is never enough to get to 0 and beyond.
Upper limit is 0.
Only one answer has 0 on RHS. That's enough.

*Lower limit? Sequence starts with $$A_1 = -\frac{1}{2}$$
But the sum of just 2 terms (let alone 200), $$A_1+A_2$$, is already > than $$-\frac{1}{2}$$
Whole sequence, summed, is thus also > than $$-\frac{1}{2}$$
Lower limit is $$-\frac{1}{2}$$
$$-\frac{1}{2} < Sum < 0$$

_________________

Never look down on anybody unless you're helping them up.
--Jesse Jackson

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

### Show Tags

05 Apr 2018, 16:10
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by $$(-1)^n*2^{(-n)}$$. If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Let’s solve the expression for values of n of 1, 2, 3, and 4.

n = 1:

-1 x 1/2 = -1/2

n = 2:

1 x 1/4 = 1/4

n = 3:

-1 x 1/8 = -1/8

n = 4:

1 x 1/16 = 1/16

Thus, the first 4 terms of the sequence are:

-1/2, 1/4, -1/8, 1/16, …

We can see that this is a geometric sequence with the first term a1 = -1/2 and the common ratio r = -1/2. Recall that the sum of a finite geometric sequence with n terms is S(n) = a1 * (1 - r^n)/(1 - r). Thus, N or S(200), is

-1/2 * (1 - (-1/2)^200)/(1 - (-1/2))

Notice that (-1/2)^200 is very small, so we can approximate it as 0. Thus, we have

-1/2 * (1 - 0)/(1 + 1/2)

(-1/2)/(3/2)

-1/3

Thus we see that N is between -1/2 and 0.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: For every integer n from 1 to 200, inclusive, the nth term of a certai &nbs [#permalink] 05 Apr 2018, 16:10
Display posts from previous: Sort by