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the first 3 terms repeat without end

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the first 3 terms repeat without end [#permalink] New post 15 May 2016, 18:14
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Question Stats:

69% (00:31) correct 30% (01:08) wrong based on 13 sessions
1, –3, 4, 1, –3, 4, 1, –3, 4,..


In the sequence above, the first 3 terms repeat without end. What is the sum of the terms of the sequence from the 150th term to the 154th term?


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Question: 18
Page: 65


[Reveal] Spoiler: OA
7

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Joined: 07 Jun 2014
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Re: the first 3 terms repeat without end [#permalink] New post 15 May 2016, 18:18
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Explanation

Examining the repeating pattern, you see that the 3rd term is 4, and every 3rd term after that, in other words, the 6th, 9th, 12th, 15th, and so on, is 4. Since 150 is a multiple of 3, the 150th term is 4. Therefore the 150th to the 154th terms are 4, 1, –3, 4, 1. The sum of these 5 terms is 7, so the correct answer is 7.
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Re: the first 3 terms repeat without end [#permalink] New post 11 May 2018, 10:48
150 is also a multiple of 2 and 5. How did you get 3 instead of 2 or 5 in this case?
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Re: the first 3 terms repeat without end [#permalink] New post 11 May 2018, 12:46
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gregregre wrote:
150 is also a multiple of 2 and 5. How did you get 3 instead of 2 or 5 in this case?


Write out the sequence:

1, –3, 4, 1, –3, 4

6th term is 4 even though 6 is divisible by 2. We divided by 3 because there are 3 different types of numbers in the list.

For example lets take the following repeating sequence:

2, 5, 7, 8, 9, 2, 5, 7, 8, 9, 2, 5, 7, 8, 9, .....

What is the 15th number? 9 remainder of(15/5)= 0

what is the 22nd number? 5 remainder of (22/5)= 2


So you can vizualize the problem as follows:



remainder 1remainder 2remainder 3remainder 4remainder 0remainder 1remainder 2remainder 3remainder 4remainder 0
2578925789


Does this help?
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Re: the first 3 terms repeat without end   [#permalink] 11 May 2018, 12:46
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