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.
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 first is twi
[#permalink]
Show Tags
15 Dec 2010, 11:08
1
7
00:00
A
B
C
D
E
Difficulty:
35% (medium)
Question Stats:
71% (01:39) correct 29% (01:40) wrong based on 299 sessions
HideShow timer Statistics
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: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi
[#permalink]
Show Tags
15 Dec 2010, 11:30
1
2
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.
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. \(2^2\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi
[#permalink]
Show Tags
03 Jan 2017, 10:11
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 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.
Pattern : 1st term, --------------------------------------, 6th term ,... can be written as : 1 ,(2) ,(2*2),(2*2*2),(2*2*2*2) , (2*2*2*2*2),...
which again can be written as : 1 , 2^1, 2^2 , 2^3 , 2^4 , 2^5 ,...
Therefore , 16th term : 2^15 ---(1) 17th term : 2^16 ---(2) 18th term : 2^17 ---(3)
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi
[#permalink]
Show Tags
15 Dec 2017, 08:07
2
Top Contributor
1
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 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.
First notice the PATTERN: term_1 = 1 (aka 2^0) term_2 = 2 (aka 2^1) term_3 = 4 (aka 2^2) term_4 = 8 (aka 2^3) term_5 = 16 (aka 2^4) . . . Notice that the exponent is 1 LESS THAN the term number.
So, term_16 = 2^15 term_17 = 2^16 term_18 = 2^17
We want to find the sum 2^15 + 2^16 + 2^17 We can do some factoring: 2^15 + 2^16 + 2^17 = 2^15(1 + 2^1 + 2^2) = 2^15(1 + 2 + 4) = 2^15(7) = E
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi
[#permalink]
Show Tags
22 Jan 2019, 06:24
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.
_________________