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

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

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Ajit

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Kudos [?]: 47463 [0], given: 7123

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.

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
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Posts: 884
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Kudos [?]: 308 [0], given: 33

Re: geometric series [#permalink]  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: 10 Feb 2011
Posts: 115
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Kudos [?]: 88 [0], given: 10

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)
Math Expert
Joined: 02 Sep 2009
Posts: 28781
Followers: 4594

Kudos [?]: 47463 [0], given: 7123

Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the f [#permalink]  14 Feb 2011, 16:33
Expert's post
Merging similar topics.
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Manager
Joined: 21 Oct 2013
Posts: 194
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
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Kudos [?]: 22 [0], given: 19

Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink]  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: 40
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Kudos [?]: 1 [0], given: 1

Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink]  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
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Kudos [?]: 7598 [1] , given: 186

Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink]  02 Jul 2014, 21:49
1
KUDOS
Expert's post
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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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the   [#permalink] 02 Jul 2014, 21:49
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