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| f(n)-f(n-1) | = | d |
|---|---|---|
| f(2)-f(1) | 7-6 | 1 |
| f(3)-f(2) | 8-7 | 1 |
| f(4)-f(3) | 9-8 | 1 |
| f(5)-f(4) | 10-9 | 1 |
As we can see, the common difference d between two consecutive terms is always 1. Therefore, it is an arithmetic sequence.
| f(n)-f(n-1) | = | d |
|---|---|---|
| f(2)-f(1) | 10-5 | 5 |
| f(3)-f(2) | 20-10 | 10 |
This time, we can stop before checking all of the differences. The difference between two consecutive terms being different even once is enough to show that it's not always the same. Therefore, 5,10,20,35,55,... is not an arithmetic sequence.
| f(n)-f(n-1) | = | d |
|---|---|---|
| f(2)-f(1) | -1-0 | -1 |
| f(3)-f(2) | 1-(-1) | 2 |
Just as in Part B, we can stop after just two calculations. We already know that the difference between consecutive terms is not always the same. Therefore, it is not an arithmetic sequence.
| f(n)-f(n-1) | = | d |
|---|---|---|
| f(2)-f(1) | 16-1 | 15 |
| f(3)-f(2) | 81-16 | 65 |
Once again, the difference between consecutive terms is not always the same and, therefore, the given sequence is not arithmetic.
| f(n)-f(n-1) | = | d |
|---|---|---|
| f(2)-f(1) | - 4-(- 2) | -2 |
| f(3)-f(2) | - 6-(- 4) | -2 |
| f(4)-f(3) | - 8-(- 6) | -2 |
| f(5)-f(4) | - 10-(- 8) | -2 |
This time, the common difference d between consecutive terms is always - 2. Therefore, this sequence is arithmetic.