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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink]
15 Dec 2010, 11:08

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

73% (02:05) correct
27% (01:34) wrong based on 102 sessions

In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twice the previous term. What is the sum of the 16th, 17th and 18th terms in the sequence ?

A. 2^18 B. 3(2^17) C. 7(2^16) D. 3(2^16) E. 7(2^15)

Could some tell me the basic formula for handling geometric series. Thanks.

Re: geometric series [#permalink]
15 Dec 2010, 11:30

Expert's post

ajit257 wrote:

In a sequence 1,2,4,8,16,32......each term after the first is twice the previous term. What is the sum of the 16th, 17th and 18th tems in the sequence ?

a. 2^18 b. 3(2^17) c. 7(2^16) d. 3(2^16) e. 7(2^15)

Could some tell me the basic formula for handling geometric series. Thanks.

So you don't actually need geometric series formula.

Answer: E.

But still if you are interested:

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 \(\neq{1}\).

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

In the sequence 1, 2, 4, 8, 16, 32, …, each term after the f [#permalink]
14 Feb 2011, 16:11

In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?

A. 2^18 B. 3(2^17) C. 7(2^16) D. 3(2^16) E. 7(2^15)

So looking at the pattern, what will be the 16th term? It will be \(2^{15}\) What about the 17th term? \(2^{16}\) What about the 18th term? \(2^{17}\)

When you add them, you get \(2^{15} + 2^{16} + 2^{17}\) Now you take \(2^{15}\) common from the 3 terms. You are left with

\(2^{15}* (1 + 2 + 2^2) = 2^{15}*7\)

Note that \(2^{16}\) has 16 2s. When you take out 15 2s, you are left with a single 2. Similarly, \(2^{17}\) has 17 2s. When you take out 15 2s, you are left with two 2s i.e. \(2^2\) _________________

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...