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

 It is currently 14 Jul 2020, 09:43 ### 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.  # The sum of the terms of a geometric progression is 2047.

Author Message
TAGS:

### Hide Tags

Current Student P
Status: Chasing my MBB Dream!
Joined: 29 Aug 2012
Posts: 1094
Location: United States (DC)
WE: General Management (Aerospace and Defense)
The sum of the terms of a geometric progression is 2047.  [#permalink]

### Show Tags

1
Top Contributor
6 00:00

Difficulty:   65% (hard)

Question Stats: 56% (01:42) correct 44% (01:37) wrong based on 116 sessions

### HideShow timer Statistics

The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

_________________
Become a GMAT Club Premium member to unlock discounts

MBA, Class of 2020,
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4961
GMAT 1: 770 Q49 V46
Re: The sum of the terms of a geometric progression is 2047.  [#permalink]

### Show Tags

2
Top Contributor
Gnpth wrote:
The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

In case there are people reading this question and worrying that they haven't learned about geometric progressions and common ratios, you need not worry. The GMAT does not expect you to be familiar with these terms. Likewise, you don't need to know the formula for the sum of a geometric series/sequence/progression.

Cheers,
Brent
_________________
VP  V
Joined: 27 May 2012
Posts: 1071
The sum of the terms of a geometric progression is 2047.  [#permalink]

### Show Tags

Gnpth wrote:
The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

Summation of n terms of a G.P = sum = a$$(\frac{ 1- r^n}{1-r})$$

Where a is the first term and n is the number of terms whose sum is being taken and r is the common ratio.

2047 = $$1 (\frac{ 1- r^n}{1-r})$$

Also nth term of a G.P.= $$ar^{(n-1)}$$ where a is the first term and r is the common ratio.
So last term ( nth term) = 1024 = $$ar^{(n-1)}$$
We know a (first term) is 1
$$1024=r^{(n-1)}$$
$$2^{10}$$ = 1024 here n-1= 10 so n=11
$$4^5$$ = 1024 here n-1= 5 so n=6
$$32^2$$ = 1024 here n-1= 2 so n=3

Out of these only n=11 will give sum of 2047 hence r= 2 is the required common ratio.

Hope this helps !
Kudos will be appreciated !
Thank you.
_________________
- Stne
GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2320
Re: The sum of the terms of a geometric progression is 2047.  [#permalink]

### Show Tags

2
As Brent says above, you don't need to know anything about "geometric series" for the GMAT - you don't even need to know what they are. So ignoring the other problems with the wording of the question, it's simply out of the scope of the test. The only way it could appear is if the question itself also provided the formulas you'd need to use to answer it.

But since the only solution above assumes r is a positive integer, and uses inspection, I can offer a different method, though test takers can safely ignore questions like this. We know from Statement 1 that the first term a = 1, and the last term is 1024. In a geometric sequence with n terms, the last term is equal to (a)(r^(n-1)). Since a =1, we know

r^(n-1) = 1024

The sum S of a geometric sequence with n terms is equal to

S = a(1- r^n)/(1 - r)

so since that sum is 2047, we also know, using that a=1,

2047 = (1 - r^n) / (1 - r)

Since r^n is just equal to (r)(r^(n-1)) by basic exponent rules, and since r^(n-1) = 1024, we can replace r^n with 1024r. So

2047 = (1 - 1024r) / (1 - r)
2047 - 2047r = 1 - 1024r
2046 = 1023r
2 = r

and Statement 1 is sufficient. Statement 2 is not sufficient (it could be the sequence with r=2 that we found in Statement 1, but since r need not be an integer, it could just be a two-term sequence where the first term is 512, and the second is 1535, among many possibilities), so the answer is A.
_________________
GMAT Tutor in Montreal

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
Intern  B
Joined: 23 Dec 2019
Posts: 3
Re: The sum of the terms of a geometric progression is 2047.  [#permalink]

### Show Tags

BrentGMATPrepNow wrote:
Gnpth wrote:
The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

In case there are people reading this question and worrying that they haven't learned about geometric progressions and common ratios, you need not worry. The GMAT does not expect you to be familiar with these terms. Likewise, you don't need to know the formula for the sum of a geometric series/sequence/progression.

Cheers,
Brent

Thank you Brent for Clarification! Re: The sum of the terms of a geometric progression is 2047.   [#permalink] 06 Jun 2020, 22:02

# The sum of the terms of a geometric progression is 2047.  