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

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Question Stats: 74% (01:31) correct 26% (01:38) wrong based on 422 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. $$7(2^{15})$$

Originally posted by ajit257 on 15 Dec 2010, 11:08.
Last edited by Bunuel on 26 Feb 2019, 03:51, edited 1 time in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 64151
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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5
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: 578
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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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: 173
Location: Germany
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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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: 31
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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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 V
Joined: 16 Oct 2010
Posts: 10485
Location: Pune, India
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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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 first is twi  [#permalink]

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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
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Joined: 11 Sep 2015
Posts: 4879
GMAT 1: 770 Q49 V46
In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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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
_________________

Originally posted by BrentGMATPrepNow on 15 Dec 2017, 08:07.
Last edited by BrentGMATPrepNow on 10 Jan 2020, 05:42, edited 1 time in total.
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In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi  [#permalink]

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Bunuel wrote:
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.

Bunuel & karishma is it important to know the 'Sum of infinite geometric progression with common ratio' for the GMAT? Many tks!  In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi   [#permalink] 14 Oct 2019, 06:58

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