GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 30 Mar 2020, 12:43 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # For every positive integer n, if the nth term of the sequence is

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 62353
For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags 00:00

Difficulty:   45% (medium)

Question Stats: 72% (02:44) correct 28% (02:49) wrong based on 71 sessions

### HideShow timer Statistics

Competition Mode Question

For every positive integer n, if the nth term of the sequence is $$x_n=\frac{1}{(n+1)(n+2)}$$ then $$x_1+x_2+x_3+x_4+x_5=$$?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Are You Up For the Challenge: 700 Level Questions

_________________
CEO  V
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3406
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

2
Quote:
For every positive integer n, if the nth term of the sequence is x_n=1/(n+1)(n+2) then x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

$$x_1 = 1/(2*3) = (1/2)-(1/3)$$
$$x_2 = 1/(3*4) = (1/3)-(1/4)$$
$$x_3 = 1/(4*5) = (1/4)-(1/5)$$
$$x_4 = 1/(5*6) = (1/5)-(1/6)$$
$$x_5 = 1/(6*7) = (1/6)-(1/7)$$

$$x_1 + x_2 + x_3 + x_4 + x_5 = (1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+(1/5)-(1/6)+(1/6)-(1/7) = (1/2)-(1/7) = 5/14$$

_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi

My Recent Posted Questions
Q-1-Integer-Divisibility l Q-2-Inequality DS l Q-3-Standard Deviation l Q-4-Inequality

ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
Senior Manager  G
Joined: 16 Feb 2015
Posts: 355
Location: United States
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
Explanation:

Xn = 1/(n+1)(n+2) = 1/(n+1) - 1/(n+2)

So,
X1 = 1/2 - 1/3
X2 = 1/3 - 1/4
X3 = 1/4 - 1/5
X4 = 1/5 - 1/6
X5 = 1/6 - 1/7

x1+x2+x3+x4+x5 = 1/2- 1/7 = 5/14

IMO-C
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 6059
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

for the given function ; find x1 to x5
we get x1 = 1/6 x2= 1/12 x3= 1/20 x4= 1/30 x5= 1/42
take LCM
sum from x1 to x5 = 150/420 ; 5/14
IMO C

For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Originally posted by Archit3110 on 30 Jan 2020, 00:51.
Last edited by Archit3110 on 31 Jan 2020, 09:22, edited 1 time in total.
Senior Manager  D
Joined: 20 Mar 2018
Posts: 492
Location: Ghana
Concentration: Finance, Real Estate
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
For every positive integer n, if the nth term of the sequence is
Xn = 1/(n+1)(n+2) then X1+X2+X3+X4+X5 =?

1/2(3)+1/3(4)+ 1/4(5) + 1/5(6)+1/6(7)
1/6 + 1/12+ 1/20 + 1/30 + 1/42
3/12+ 1/20 +1/30+ 1/42
15/60+ 3/60 + 2/60 +1/42
1/3 + 1/42 —> 14/42 + 1/42 = 5/14

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Hit that C

Posted from my mobile device
Director  V
Joined: 30 Sep 2017
Posts: 804
GMAT 1: 720 Q49 V40 GPA: 3.8
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
x1+x2+x3+x4+x5
=1/6+1/12+1/20+1/30+1/42
=1/420*(70+35+21+14+10)
=150/420
=5/14

Posted from my mobile device
VP  P
Joined: 24 Nov 2016
Posts: 1353
Location: United States
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
Quote:
For every positive integer n, if the nth term of the sequence is x_n=1/(n+1)(n+2) then x1+x2+x3+x4+x5= ?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

x_n=1/(n+1)(n+2)
x_1=1/2*3, x_2=1/3*4, x_3=1/4*5, x_4=1/5*6, x_5=1/6*7
lcm(3,4,5,7)=420: sum(x_1 to x_5)=(70+35+21+14+10)/420=5/14

OR
[1+2] 1/2*3+1/3*4=2+4/2*3*4=1/4
[3+4] 1/4*5+1/5*6=6+4/4*5*6=10/120=1/12
[1,2,3,4] 1/4+1/12=3+1/4*3=4/12=1/3
[(1,2,3,4)+5]=1/3+1/6*7=1/3+1/3*14=15/42=5/14

Ans (C)

Originally posted by exc4libur on 30 Jan 2020, 03:30.
Last edited by exc4libur on 30 Jan 2020, 03:37, edited 1 time in total.
VP  V
Joined: 20 Jul 2017
Posts: 1466
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
Given, $$x_n = \frac{1}{(n+1)(n+2)}$$
--> $$x_1$$ + $$x_2$$ + $$x_3$$ + $$x_4$$ + $$x_5$$ = $$\frac{1}{(2)(3)} + \frac{1}{(3)(4)} + \frac{1}{(4)(5)} + \frac{1}{(5)(6)} + \frac{1}{(6)(7)}$$
= $$\frac{3 - 2}{2*3} + \frac{4 - 3}{3*4} + \frac{5 - 4}{4*5} + \frac{6 - 5}{5*6} + \frac{7 - 6}{6*7}$$
= $$\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}$$
= $$\frac{1}{2} - \frac{1}{7}$$
= $$\frac{7 - 2}{2*7}$$
= $$\frac{5}{14}$$

Option C
CR Forum Moderator P
Joined: 18 May 2019
Posts: 807
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
x1=1/(2*3)=1/6
x2=1/(3*4)=1/12
x3=1/(4*5)=1/20
x4=1/(5*6)=1/30
x5=1/(6*7)=1/42

x1+x2+x3+x4+x5 = 1/6+1/12+1/20+1/30+1/42 =[70+35+21+14+10]/420 = 150/420 = 15/42 = 5/14

Posted from my mobile device
Intern  B
Joined: 13 Jan 2020
Posts: 18
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
$$x_n = \frac{1}{{(n+1)(n+2)}}$$

$$x_1 + x_2 + x_3 + x_4 + x_5 = \frac{1}{{(2)(3)}} + \frac{1}{{(3)(4)}} + \frac{1}{{(4)(5)}} + \frac{1}{{(5)(6)}} +\frac{ 1}{{(6)(7)}}$$

$$= \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + \frac{1}{4} -\frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}$$
$$= \frac{1}{2} - \frac{1}{7}$$
$$= \frac{(7 - 2)}{(2*7)}$$
$$= \frac{5}{14 }$$
Senior Manager  G
Joined: 14 Dec 2019
Posts: 475
Location: Poland
GMAT 1: 570 Q41 V27
WE: Engineering (Consumer Electronics)
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

x1 = $$1/(2*3)$$
x2 = $$1/(3*4)$$

x1 + x2 = $$1/4$$

x3 = $$1/(4*5)$$
x4 = $$1/(5*6)$$

x3 + x4 = $$1/12$$

x5 = $$1/42$$

Therefore x1 + x2 + x3 + x4 + x5 = $$1/4+1/12+1/42$$ -> $$(21+7+2)/84$$ -> $$5/14$$ - Answer C
Director  P
Joined: 25 Jul 2018
Posts: 647
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
For every positive integer n, if the nth term of the sequence is $$x_n=1/(n+1)(n+2)$$ then $$x_1+x_2+x_3+x_4+x_5$$=?

--> $$\frac{1}{2*3}+ \frac{1}{3*4}+\frac{1}{4*5}+\frac{1}{5*6}+ \frac{1}{6*7} =$$

$$= (\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7}) =$$

$$= \frac{1}{2} -\frac{1}{7}= \frac{(7-2)}{2*7}= \frac{5}{14}$$

CEO  V
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3406
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

Archit3110 wrote:
for the given function ; find x1 to x5
we get x1 = 1/6 x2= 1/12 x3= 1/20 x4= 1/30 x5= 1/58
sum from x1 to x5 = 75/240 ; 5/16
IMO D

For every positive integer n, if the nth term of the sequence is xn=1(n+1)(n+2)xn=1(n+1)(n+2) then x1+x2+x3+x4+x5=x1+x2+x3+x4+x5=?

A. 4/5
B. 2/3
C. 5/14
D. 5/16
E. 6/19

Archit3110
Some typo in highlighted part ... should be 1/42 _________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi

My Recent Posted Questions
Q-1-Integer-Divisibility l Q-2-Inequality DS l Q-3-Standard Deviation l Q-4-Inequality

ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
Intern  Joined: 20 Oct 2019
Posts: 4
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

1
Hey! Could someone please explain how is the formula 1/(x+1)(x+2) derived into 1/(x+1) - 1/(x-2)? Why do we subtract??
Intern  B
Joined: 27 Apr 2018
Posts: 1
Location: Mexico
Re: For every positive integer n, if the nth term of the sequence is  [#permalink]

### Show Tags

IevaPieva wrote:
Hey! Could someone please explain how is the formula 1/(x+1)(x+2) derived into 1/(x+1) - 1/(x-2)? Why do we subtract??

I was dealing with the same question to improve the way in which this type of question is solved. Although I cannot explain the theory behind this reasoning, I would try to show an example:

We have that --> xn = 1/(x+1)(x+2) and therefore, for x1 = 1 / (1+1)(1+2) --> 1 / (2)(3) --> 1/6.

It turns that another way to obtain 1/6 is if we separate both fractions (as in the solution provided by the other members). x1 = 1 / (1+1) - 1 /(1+2) --> 1/2 - 1/3 --> 3-2/(2)(3). That's why there is a subtraction. Re: For every positive integer n, if the nth term of the sequence is   [#permalink] 21 Feb 2020, 11:14
Display posts from previous: Sort by

# For every positive integer n, if the nth term of the sequence is  