Find the first four terms and state whether the sequence is arithmetic, geometric, or neither.
 a_{n} = 3n + 2
To find the first four terms, in a row, replace n with 1, 2, 3 and 4
3n + 2 = 3(1) + 2 = 3 + 2 = 5
3n + 2 = 3(2) + 2 = 6 + 2 = 8
3n + 2 = 3(3) + 2 = 9 + 2 = 11
3n + 2 = 3(4) + 2 = 12 + 2 = 14
Answer: The first four terms of the sequence are 5, 8, 11, 14. This sequence is arithmetic with common difference of d = 3
 a_{n} = n^{2} + 1
substituting n = 1,2,3,4
n^{2} + 1 = 1^{2} + 1 = 1 + 1 = 2
n^{2} + 1 = 2^{2} + 1 = 4 + 1 = 5
n^{2} + 1 = 3^{2} + 1 = 9 + 1 = 10
n^{2} + 1 = 4^{2} + 1 = 16 + 1 = 17
Answer: The first four terms of the sequence are 2, 5, 10, 17. The sequence is neither arithmetic nor geometric.
 a_{n} = 3.2^{n}
substituting n = 1,2,3,4
3.2^{n} = 3.2^{1} = 3 x 2 = 6
3.2^{n} = 3.2^{2} = 3 x 4 = 12
3.2^{n} = 3.2^{3} = 3 x 8 = 24
3.2^{n} = 3.2^{4} = 3 x 16 = 48
Answer: The first four terms of the sequence are 6, 12, 24, 48. The sequence is geometric with r = 2.
Directions: Answer the following questions. Also write at least 5 examples of your own.
