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A sequence of numbers is given as: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 and so
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13 May 2024, 03:03
13 May 2024, 03:03
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Expert Reply
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Question Stats:
86% (01:58) correct
13% (04:09) wrong based on 15 sessions
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A sequence of numbers is given as: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 and so on. What will be the 100th term of this sequence?
A. 11
B. 12
C. 13
D. 14
E. 15
A. 11
B. 12
C. 13
D. 14
E. 15
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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
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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: A sequence of numbers is given as: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 and so
[#permalink]
19 May 2024, 17:21
19 May 2024, 17:21
Is there a way to solve this problem quicker besides writing out the terms?
A sequence of numbers is given as: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 and so
[#permalink]
19 May 2024, 23:02
19 May 2024, 23:02
Expert Reply
Of course . Using observation and logic.
Notice how the sequence has as first term 1 then 2 but two times and then three three times , 4 four times............until k which is the 100 terms
This mean that using the formula for sum of an arithmetic progression we can find easily the solution
\(\frac{k(k+1)}{2}\) why ? every term after the previous June is oNE more
Picking number for example C we do have 13*14/2=91 and the answer is D or E
We need a value a bit bigger (100) so the answer is D
Notice how the sequence has as first term 1 then 2 but two times and then three three times , 4 four times............until k which is the 100 terms
This mean that using the formula for sum of an arithmetic progression we can find easily the solution
\(\frac{k(k+1)}{2}\) why ? every term after the previous June is oNE more
Picking number for example C we do have 13*14/2=91 and the answer is D or E
We need a value a bit bigger (100) so the answer is D
