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Sequence S is such that Sn
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Updated on: 19 Mar 2019, 04:55
Updated on: 19 Mar 2019, 04:55
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80% (02:35) correct
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Sequence S is such that \(S_n = S_{n-1} + \frac{3}{2}\) and \(S_1 = 2\)
Sequence A is such that \(A_n = A_{n-1} - 1.5\) and \(A_1 = 18.5\)
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Sequence A is such that \(A_n = A_{n-1} - 1.5\) and \(A_1 = 18.5\)
Quantity A |
Quantity B |
The sum of the terms in S from \(S_1\) to \(S_{13}\), inclusive |
The sum of the terms from in A from \(A_1\) to \(A_{13}\), inclusive |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Originally posted by kaka on 13 Mar 2019, 00:34.
Last edited by GreenlightTestPrep on 19 Mar 2019, 04:55, edited 3 times in total.
Last edited by GreenlightTestPrep on 19 Mar 2019, 04:55, edited 3 times in total.
Edited by Carcass
Re: Sequence S is such that Sn
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13 Mar 2019, 01:15
13 Mar 2019, 01:15
Expert Reply
Please kaka format the question properly. It is really easy to do
https://gre.myprepclub.com/forum/qq-how-to ... -2357.html
Thank you so much
Are you sure the terms in the question are correct ? I did find an equal question but different values.
Let me know
https://gre.myprepclub.com/forum/qq-how-to ... -2357.html
Thank you so much
Are you sure the terms in the question are correct ? I did find an equal question but different values.
Let me know
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Joined: 10 Apr 2015
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Given Kudos: 136
Re: Sequence S is such that Sn
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19 Mar 2019, 04:55
19 Mar 2019, 04:55
1
kaka wrote:
Sequence S is such that \(S_n = S_{n-1} + \frac{3}{2}\) and \(S_1 = 2\)
Sequence A is such that \(A_n = A_{n-1} - 1.5\) and \(A_1 = 18.5\)
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Sequence A is such that \(A_n = A_{n-1} - 1.5\) and \(A_1 = 18.5\)
Quantity A |
Quantity B |
The sum of the terms in S from \(S_1\) to \(S_{13}\), inclusive |
The sum of the terms from in A from \(A_1\) to \(A_{13}\), inclusive |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
The first 13 terms of SEQUENCE S: 2, 3.5, 5, 6.5, 8, 9.5, 11, 12.5, 14, 15.5, 17, 18.5, 20
The first 13 terms of SEQUENCE A: 18.5, 17, 15.5, 14, 12.5, 11, 9.5, 8, 6.5, 5, 3.5, 2, 0.5
We can see that the two sequence share 12 terms
If we subtract those 12 shared terms from each sequence, we get:
The first 13 terms of SEQUENCE S:
The first 13 terms of SEQUENCE A:
We can see that Quantity A is greater.
Answer: A
Cheers,
Brent
