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In a sequence a1, a2, …, each term is defined as an = 1/2^n. Which of
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13 May 2021, 10:11
13 May 2021, 10:11
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In a sequence \(a_1\), \(a_2\), …, each term is defined as \(a_n=\frac{1}{2^n}\). Which of the following expressions represents the sum of the first 10 terms of \(a_n\) ?
(A) \(1-\frac{1}{2^{10}}\)
(B) \(1- \frac{1}{2^9}\)
(C) \(1+\frac{1}{2^9}\)
(D) \(1+\frac{1}{2^{10}}\)
(E) \(1+\frac{1}{2^{11}}\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) \(1-\frac{1}{2^{10}}\)
(B) \(1- \frac{1}{2^9}\)
(C) \(1+\frac{1}{2^9}\)
(D) \(1+\frac{1}{2^{10}}\)
(E) \(1+\frac{1}{2^{11}}\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: In a sequence a1, a2, …, each term is defined as an = 1/2^n. Which of
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13 May 2021, 10:27
13 May 2021, 10:27
1
Carcass wrote:
In a sequence \(a_1\), \(a_2\), …, each term is defined as \(a_n=\frac{1}{2^n}\). Which of the following expressions represents the sum of the first 10 terms of \(a_n\) ?
(A) \(1-\frac{1}{2^{10}}\)
(B) \(1- \frac{1}{2^9}\)
(C) \(1+\frac{1}{2^9}\)
(D) \(1+\frac{1}{2^{10}}\)
(E) \(1+\frac{1}{2^{11}}\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) \(1-\frac{1}{2^{10}}\)
(B) \(1- \frac{1}{2^9}\)
(C) \(1+\frac{1}{2^9}\)
(D) \(1+\frac{1}{2^{10}}\)
(E) \(1+\frac{1}{2^{11}}\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Let's look for a pattern
If \(n=1\), the sequence is \(\frac{1}{2}\). So the SUM = \(\frac{1}{2}\)
If \(n=2\), the sequence is \(\frac{1}{2}\), \(\frac{1}{4}\). So the SUM = \(\frac{3}{4}\)
If \(n=3\), the sequence is \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\). So the SUM = \(\frac{7}{8}\)
If \(n=4\), the sequence is \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\). So the SUM = \(\frac{15}{16}\)
At this point, we may notice that the SUM is always less than 1.
So, we can ELIMINATE C, D and E
Now examine the remaining answer choices (A and B).
This gives us a hint about the correct answer.
Notice that, when we have \(n\) terms, the SUM is always \(\frac{1}{2^n}\) LESS THAN 1
The last term in the given sequence is \(\frac{1}{2^{10}}\)
So, the sum will equal \(1-\frac{1}{2^{10}}\)
Answer: A
Cheers,
Brent
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In a sequence a1, a2, …, each term is defined as an = 1/2^n. Which of
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08 Jul 2021, 04:45
08 Jul 2021, 04:45
Theory: Geometric Series is a series in which consecutive terms have the same ratio.
Sum of n terms of the Geometric series is given by \(S{_n} = a* \frac{(1 - r^n) }{ (1-r)}\) for r < 1
Where, a is the first term
r is the common ratio (or \(\frac{T{_2} }{ T{_1}}\) )
n is the term number
Now, we have a series \(a_1\), \(a_2\), …, where each term is defined as \(a_n=\frac{1}{2^n}\)
So, \(a_1\) = \(\frac{1}{2^1}\) = \(\frac{1}{2}\)
\(a_2\) = \(\frac{1}{2^2}\) = \(\frac{1}{4}\) and so on.
So, the series is \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\) and so on...
So, First term, a = \(\frac{1}{2}\)
Common ratio, r = \(\frac{1}{4}\) / \(\frac{1}{2}\) = \(\frac{1}{2}\)
So, Sum of first 10 terms \(S_10\) is given by
\(S_10 = a* \frac{(1 - r^{10}) }{ (1-r)}\) [ Putting n = 1- in \(S{_n} = a* \frac{(1 - r^n) }{ (1-r)}\) ]
=> \(S_{10}\) = \(\frac{1}{2}* (1 - \frac{1}{2}^{10}) / (1-\frac{1}{2})\)
= \(\frac{1}{2}* (1 - \frac{1}{2}^{10}) / (\frac{1}{2})\)
= \(1 - \frac{1}{2}^{10}\)
So, answer will be A
Hope it helps!
To learn more about the Geometric and Arithmetic series watch the below video
Sum of n terms of the Geometric series is given by \(S{_n} = a* \frac{(1 - r^n) }{ (1-r)}\) for r < 1
Where, a is the first term
r is the common ratio (or \(\frac{T{_2} }{ T{_1}}\) )
n is the term number
Now, we have a series \(a_1\), \(a_2\), …, where each term is defined as \(a_n=\frac{1}{2^n}\)
So, \(a_1\) = \(\frac{1}{2^1}\) = \(\frac{1}{2}\)
\(a_2\) = \(\frac{1}{2^2}\) = \(\frac{1}{4}\) and so on.
So, the series is \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\) and so on...
So, First term, a = \(\frac{1}{2}\)
Common ratio, r = \(\frac{1}{4}\) / \(\frac{1}{2}\) = \(\frac{1}{2}\)
So, Sum of first 10 terms \(S_10\) is given by
\(S_10 = a* \frac{(1 - r^{10}) }{ (1-r)}\) [ Putting n = 1- in \(S{_n} = a* \frac{(1 - r^n) }{ (1-r)}\) ]
=> \(S_{10}\) = \(\frac{1}{2}* (1 - \frac{1}{2}^{10}) / (1-\frac{1}{2})\)
= \(\frac{1}{2}* (1 - \frac{1}{2}^{10}) / (\frac{1}{2})\)
= \(1 - \frac{1}{2}^{10}\)
So, answer will be A
Hope it helps!
To learn more about the Geometric and Arithmetic series watch the below video
