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

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

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

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.

Hope it helps.
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Re: geometric series [#permalink] New post 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.
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In the sequence 1, 2, 4, 8, 16, 32, …, each term after the f [#permalink] New post 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)
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the f [#permalink] New post 14 Feb 2011, 16:33
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink] New post 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.
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink] New post 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
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink] New post 02 Jul 2014, 21:49
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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   [#permalink] 02 Jul 2014, 21:49
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