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the first 3 terms repeat without end
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15 May 2016, 18:14
15 May 2016, 18:14
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Question Stats:
61% (01:17) correct
38% (01:16) wrong based on 131 sessions
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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?
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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Re: the first 3 terms repeat without end
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11 May 2018, 12:46
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 1 | remainder 2 | remainder 3 | remainder 4 | remainder 0 | remainder 1 | remainder 2 | remainder 3 | remainder 4 | remainder 0 |
| 2 | 5 | 7 | 8 | 9 | 2 | 5 | 7 | 8 | 9 |
Does this help?
General Discussion
Re: the first 3 terms repeat without end
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15 May 2016, 18:18
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.
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.
