How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 14:14

### 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 June
Open Detailed Calendar

# How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9

Author Message
TAGS:

### Hide Tags

Manager
Status: Taking heavily leveraged but calculated risks at all times
Joined: 04 Apr 2010
Posts: 185
Concentration: Entrepreneurship, Finance
Schools: HBS '15, Stanford '15
GMAT Date: 01-31-2012
Followers: 1

Kudos [?]: 68 [3] , given: 12

How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

08 Dec 2011, 09:03
3
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

55% (03:10) correct 45% (06:34) wrong based on 334 sessions

### HideShow timer Statistics

How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 + ... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9
[Reveal] Spoiler: OA
Manager
Joined: 24 Oct 2011
Posts: 95
Location: India
GMAT Date: 11-29-2011
GPA: 3.5
WE: Web Development (Computer Software)
Followers: 1

Kudos [?]: 16 [3] , given: 27

### Show Tags

08 Dec 2011, 09:19
3
KUDOS
divide series in rational and irrational no's since sum of rational is always rational and irrational is always irrational that means in the above series
120=3+9+27+81....4 terms
121*(3^1/2)=(1+3+9+27+81)*(3^1/2).... 5 terms
so ans is 9....E
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1183
Followers: 422

Kudos [?]: 1512 [0], given: 4

### Show Tags

08 Dec 2011, 14:46
avenkatesh007 wrote:
divide series in rational and irrational no's since sum of rational is always rational and irrational is always irrational

That's true in this question, but not in general. The sum of two rational numbers is always rational, but the sum of two irrational numbers does not need to be irrational. For example, √2 and 2 - √2 are both irrational, but their sum is 2, a rational number. That's not the kind of thing the GMAT ever tests, however.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Manager
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 32 [0], given: 17

### Show Tags

23 Apr 2012, 10:41
I did not understand the above explanation. Can someone please explain?
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93660 [6] , given: 10583

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

24 Apr 2012, 12:49
6
KUDOS
Expert's post
2
This post was
BOOKMARKED
Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9

We have a following sequence: $$\sqrt{3}$$, $$3$$, $$3\sqrt{3}$$, $$9$$, ...

Notice that this sequence is a geometric progression with first term equal to $$\sqrt{3}$$ and common ratio also equal to $$\sqrt{3}$$.

The question asks: the sum of how many terms of this sequence adds up to $$120+121\sqrt{3}$$.

The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, (where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$).

So, $$\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}$$ --> $$\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n=243$$ --> $$3^{\frac{1+n}{2}}=3^5$$ --> $$\frac{1+n}{2}=5$$ --> $$n=9$$.

Shortcut solution:
Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: $$\frac{3(3^k-1)}{3-1}=120$$ --> $$3^k-1=80$$ --> $$3^k=81$$ --> $$k=4$$, so the number of terms must be either $$2k=8$$ (if there are equal number of irrational and rational terms) or $$2k+1=9$$ (if # of irrational terms, with $$\sqrt{3}$$, is one more than # rational terms), since only 9 is present among options then it must be a correct answers.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93660 [0], given: 10583

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

04 Jun 2013, 04:54
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7132
Location: Pune, India
Followers: 2140

Kudos [?]: 13715 [1] , given: 222

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

04 Jun 2013, 21:03
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9

In a series question, it is always a good idea to write the first few terms to figure out what the series looks like:

r = $$\sqrt{3} + 3 + 3\sqrt{3} + 9 + 9\sqrt{3} + 27 + 27\sqrt{3} + ....$$

Sum of first 2 terms$$= 3 + \sqrt{3}$$
Sum of first 3 terms $$= 3 + (\sqrt{3} + 3\sqrt{3})$$
Sum of first 4 terms$$= (3+9) + 4\sqrt{3}$$
Sum of first 5 terms $$= 12 + (4 + 9)\sqrt{3}$$
etc

You want a sum of $$120 + 121\sqrt{3}$$

You will obtain 120 by adding 3 + 9 + 27 + 81 (4 terms) and $$121\sqrt{3}$$ by adding $$(1 + 3 + 9 + 27 + 81)\sqrt{3}$$ (i.e. 5 terms)

So you need to add a total of 9 terms.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 28 Feb 2012
Posts: 115
GPA: 3.9
WE: Marketing (Other)
Followers: 0

Kudos [?]: 42 [0], given: 17

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

05 Jun 2013, 06:26
My method is not very sophisticated and most probably not the best but in some cases it works.
The most crucial part here is to understand the question. Basically we have a sequesnce \sqrt{3}+3+3\sqrt{3}+ .... which is the same as: \sqrt{3}+\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}*\sqrt{3}+....

As it could be seen each time next term is multiplied by \sqrt{3}.

Now the questions asks us to find how many numbers this sequence should consists of in order to be equal of 120 + 121*\sqrt{3}.

What i did, i have just tried to make the calculation easier and tried to simplify the 120 + 121*\sqrt{3}=40*\sqrt{3}*\sqrt{3}+121*\sqrt{3}=\sqrt{3}(40\sqrt{3}+121)

Now go back to our sequence, if we take the common \sqrt{3} out of the brackets we have \sqrt{3}(1+\sqrt{3}+3+3\sqrt{3}+9+9\sqrt{3}) ---> \sqrt{3}(13+13\sqrt{3}) This is the sum of 6 terms in the sequence, which is clearly less than \sqrt{3}(40\sqrt{3}+121). We can go and check whether 9 (next possible choice) will fit, but considering that 9 is the only choice left it should be it. The answer is E.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93660 [0], given: 10583

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

05 Jun 2013, 07:42
ziko wrote:
My method is not very sophisticated and most probably not the best but in some cases it works.
The most crucial part here is to understand the question. Basically we have a sequesnce \sqrt{3}+3+3\sqrt{3}+ .... which is the same as: \sqrt{3}+\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}+\sqrt{3}*\sqrt{3}*\sqrt{3}*\sqrt{3}+....

As it could be seen each time next term is multiplied by \sqrt{3}.

Now the questions asks us to find how many numbers this sequence should consists of in order to be equal of 120 + 121*\sqrt{3}.

What i did, i have just tried to make the calculation easier and tried to simplify the 120 + 121*\sqrt{3}=40*\sqrt{3}*\sqrt{3}+121*\sqrt{3}=\sqrt{3}(40\sqrt{3}+121)

Now go back to our sequence, if we take the common \sqrt{3} out of the brackets we have \sqrt{3}(1+\sqrt{3}+3+3\sqrt{3}+9+9\sqrt{3}) ---> \sqrt{3}(13+13\sqrt{3}) This is the sum of 6 terms in the sequence, which is clearly less than \sqrt{3}(40\sqrt{3}+121). We can go and check whether 9 (next possible choice) will fit, but considering that 9 is the only choice left it should be it. The answer is E.

_________________
Manager
Joined: 26 May 2012
Posts: 50
Concentration: Marketing, Statistics
Followers: 0

Kudos [?]: 8 [0], given: 11

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

11 Mar 2014, 21:31
Is the original question missing a "+9" in the portion describing the sequence? It was hard to know what the series would with just 3 terms (as the question currently is). For example, if we only know 3^(1/2), 3, 3*3^(1/2), one could interpret that the series is defined as every input starting from the 3rd is the product of the previous 2. Hope this makes sense. Please edit the question stem because I see a +9 in the answer explanations but not the question itself.
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93660 [0], given: 10583

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.. [#permalink]

### Show Tags

11 Mar 2014, 22:18
catalysis wrote:
Is the original question missing a "+9" in the portion describing the sequence? It was hard to know what the series would with just 3 terms (as the question currently is). For example, if we only know 3^(1/2), 3, 3*3^(1/2), one could interpret that the series is defined as every input starting from the 3rd is the product of the previous 2. Hope this makes sense. Please edit the question stem because I see a +9 in the answer explanations but not the question itself.

________
Edited. Thank you.
_________________
Intern
Joined: 20 May 2014
Posts: 39
Followers: 0

Kudos [?]: 5 [0], given: 1

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

03 Jul 2014, 12:01
Bunuel wrote:
Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9

We have a following sequence: $$\sqrt{3}$$, $$3$$, $$3\sqrt{3}$$, $$9$$, ...

Notice that this sequence is a geometric progression with first term equal to $$\sqrt{3}$$ and common ratio also equal to $$\sqrt{3}$$.

The question asks: the sum of how many terms of this sequence adds up to $$120+121\sqrt{3}$$.

The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, (where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$).

So, $$\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}$$ --> $$\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n=243$$ --> $$3^{\frac{1+n}{2}}=3^5$$ --> $$\frac{1+n}{2}=5$$ --> $$n=9$$.

Shortcut solution:
Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: $$\frac{3(3^k-1)}{3-1}=120$$ --> $$3^k-1=80$$ --> $$3^k=81$$ --> $$k=4$$, so the number of terms must be either $$2k=8$$ (if there are equal number of irrational and rational terms) or $$2k+1=9$$ (if # of irrational terms, with $$\sqrt{3}$$, is one more than # rational terms), since only 9 is present among options then it must be a correct answers.

Hope it's clear.

I got confused on how you got from $$\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}$$ --> $$\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}$$ to --> $$\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n=243$$ --> $$3^{\frac{1+n}{2}}=3^5$$ --> $$\frac{1+n}{2}=5$$ --> $$n=9$$.

So the part after the cross multiplication, can you break it down further for me? Like what happened to the 121 *3 ?
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93660 [0], given: 10583

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

03 Jul 2014, 12:31
sagnik2422 wrote:
Bunuel wrote:
Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9

We have a following sequence: $$\sqrt{3}$$, $$3$$, $$3\sqrt{3}$$, $$9$$, ...

Notice that this sequence is a geometric progression with first term equal to $$\sqrt{3}$$ and common ratio also equal to $$\sqrt{3}$$.

The question asks: the sum of how many terms of this sequence adds up to $$120+121\sqrt{3}$$.

The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, (where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$).

So, $$\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}$$ --> $$\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n=243$$ --> $$3^{\frac{1+n}{2}}=3^5$$ --> $$\frac{1+n}{2}=5$$ --> $$n=9$$.

Shortcut solution:
Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: $$\frac{3(3^k-1)}{3-1}=120$$ --> $$3^k-1=80$$ --> $$3^k=81$$ --> $$k=4$$, so the number of terms must be either $$2k=8$$ (if there are equal number of irrational and rational terms) or $$2k+1=9$$ (if # of irrational terms, with $$\sqrt{3}$$, is one more than # rational terms), since only 9 is present among options then it must be a correct answers.

Hope it's clear.

I got confused on how you got from $$\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}$$ --> $$\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}$$ to --> $$\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}$$ --> $$\sqrt{3}*\sqrt{3}^n=243$$ --> $$3^{\frac{1+n}{2}}=3^5$$ --> $$\frac{1+n}{2}=5$$ --> $$n=9$$.

So the part after the cross multiplication, can you break it down further for me? Like what happened to the 121 *3 ?

$$120\sqrt{3}+121*3-120-121\sqrt{3}$$;
$$(121*3-120)+(120\sqrt{3}-121\sqrt{3})$$;
$$243-\sqrt{3}$$.

$$\sqrt{3}*\sqrt{3}^n=243$$;
$$3^{\frac{1}{2}}*3^{\frac{n}{2}}=3^5$$;
$$3^{\frac{1+n}{2}}=3^5$$;
$$\frac{1+n}{2}=5$$;
$$n=9$$.

Hope it's clear now.
_________________
Intern
Joined: 25 Aug 2014
Posts: 3
Location: India
GMAT 1: 710 Q47 V39
GPA: 2.95
WE: Military Officer (Military & Defense)
Followers: 0

Kudos [?]: 6 [0], given: 1

How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

10 Sep 2014, 02:10
Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 + ... will form 120 + 121*(3^(1/2))?

A. 3
B. 4
C. 5
D. 6
E. 9

Since four of the of the terms are provided and is = 12 +4*3^(1/2) approx 18, A & B eliminated
add 9*3^(1/2) approx 15 so sum approx 33 eliminated
once more and D eliminated , leaving E as the correct answer
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13547
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

26 Oct 2015, 13:43
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13547
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

17 Jan 2017, 19:18
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.
_________________
Intern
Joined: 04 Sep 2014
Posts: 4
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9 [#permalink]

### Show Tags

17 Jan 2017, 20:29
For me, as soon as I see that 1+3+27+81 = 120, x1 +3

I know the the series is either of the form: ? + 3 + ? + 9 + ? + 27 + ? + 81 + ? (9 terms)
or
? + 3 + ? + 9 + ? + 27 + ? + 81 (8 terms)...
Re: How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9   [#permalink] 17 Jan 2017, 20:29
Similar topics Replies Last post
Similar
Topics:
Given the two equations 3r + s = 17 and r + 2s = 9, by how much does 1 07 Feb 2016, 09:04
how many minimum number of terms should be considered for 2 26 Jul 2014, 22:19
39 If (10^4 * 3.456789)^8 is written as a single term, how many 9 11 Mar 2013, 22:19
15 r=3^(n+2), then in terms of r, 9^n= 5 19 Feb 2011, 07:47
7 Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)..... 6 06 Nov 2009, 14:30
Display posts from previous: Sort by