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# In a sequence 1, 2, 4, 8, 16, 32, ... each term after the

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Manager
Joined: 28 Aug 2010
Posts: 184
In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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15 Dec 2010, 11:08
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Difficulty:

35% (medium)

Question Stats:

70% (01:39) correct 30% (01:43) wrong based on 231 sessions

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 50572

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15 Dec 2010, 11:30
1
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 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.

Given:
$$a_1=2^0=1$$;
$$a_2=2^1=2$$;
$$a_3=2^2=4$$;
...
$$a_n=2^{n-1}$$;

Thus $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}$$.

So you don't actually need geometric series formula.

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.

Hope it helps.
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Director
Joined: 03 Sep 2006
Posts: 794

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15 Dec 2010, 20:21
Given:
$$a_1=2^0=1$$;
$$a_2=2^1=2$$;
$$a_3=2^2=4$$;
...
$$a_n=2^{n-1}$$;

Thus $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}$$.

So you don't actually need geometric series formula.

Thanks very Much! This is an excellent approach.
Manager
Joined: 21 Oct 2013
Posts: 187
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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18 Jun 2014, 03:23
16th term = 2^15 (since 2^0 = 1). Hence we need 2^15+2^16+2^17.

Now take smaller numbers: 2²+2³+2^4 = 28 = 7*(2²) (which is the first term), hence 7*(2^15) will be right. E.
Intern
Joined: 20 May 2014
Posts: 32
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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02 Jul 2014, 17:54
I don't understand where the 2^16 and 2^17 go. and why is a16 + a17 + a18 = 2^15 + 2^16 + 2^17

Note : Sorry I can't do the subscripts for the a's
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8527
Location: Pune, India
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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02 Jul 2014, 21:49
1
sagnik2422 wrote:
I don't understand where the 2^16 and 2^17 go. and why is a16 + a17 + a18 = 2^15 + 2^16 + 2^17

Note : Sorry I can't do the subscripts for the a's

1st term: $$1 = 2^0$$
2nd term: $$2^1$$
3rd term: $$2^2$$
4th term: $$2^3$$
5th term: $$2^4$$

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. $$2^2$$
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Intern
Joined: 22 Jul 2016
Posts: 23
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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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)

2^15 + 2^16 + 2^17 = 2^15(1+ 2^1 + 2^2) = 2^15 ( 1+2+4) = 2^15 (7)

Ans : E
CEO
Joined: 11 Sep 2015
Posts: 3113
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the  [#permalink]

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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

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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the &nbs [#permalink] 15 Dec 2017, 08:07
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# In a sequence 1, 2, 4, 8, 16, 32, ... each term after the

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