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General term in a sequence is given as A_k=1/k
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20 May 2024, 00:35
20 May 2024, 00:35
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Question Stats:
66% (01:25) correct
33% (02:34) wrong based on 9 sessions
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General term in a sequence is given as \(A_k=\frac{1}{k}-\frac{1}{k+1}\), where \(k =2,3,4......,100\). What is the sum of all terms in this series ?
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{97}{202}\)
D. \(\frac{99}{202}\)
E. 1
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{97}{202}\)
D. \(\frac{99}{202}\)
E. 1
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Free Materials for the GRE General Exam - Where to get it!!
Shorter GRE - Preparing for the Quantitative Reasoning Measure
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Free Materials for the GRE General Exam - Where to get it!!
Shorter GRE - Preparing for the Quantitative Reasoning Measure
Shorter GRE - Preparing for the Verbal Reasoning Measure
Re: General term in a sequence is given as A_k=1/k
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20 May 2024, 13:01
20 May 2024, 13:01
1
Hey, what's up, great question.
This is a pattern recognition one, which is easy if you don't add each term but rather write them out. For example the first term: 1/2 -1/3 = 1/6, instead just write 1/2 -1/3 and move to the next term which would be 1/3-1/4 which is equal to: 1/2- 1/3+ 1/3- 1/4 you can now eliminate the 1/3. The pattern is all the middle will be canceled leaving you with the first and last term 1/2- 1/101=99/202.
This is a pattern recognition one, which is easy if you don't add each term but rather write them out. For example the first term: 1/2 -1/3 = 1/6, instead just write 1/2 -1/3 and move to the next term which would be 1/3-1/4 which is equal to: 1/2- 1/3+ 1/3- 1/4 you can now eliminate the 1/3. The pattern is all the middle will be canceled leaving you with the first and last term 1/2- 1/101=99/202.
gmatclubot
Re: General term in a sequence is given as A_k=1/k [#permalink]
20 May 2024, 13:01
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