The number of terms n can be calculated using the formula: n = (last - first) / difference + 1. Here, n = (50 - 10) / 5 + 1 = 9.

The nth term of an arithmetic sequence is given by a + (n-1)d. Here, a = 5, d = 3, n = 10. So, 5 + (10-1)3 = 32.

Substituting n = 7 into the formula gives a_7 = 5*7 - 3 = 35 - 3 = 32.

The constant ratio of consecutive terms in a geometric series is called the common ratio.

The common difference can be found by calculating S_n - S_(n-1). Here, S_n = 3n^2 + 2n and S_(n-1) = 3(n-1)^2 + 2(n-1). The difference simplifies to 4.

'a' represents the first term of the arithmetic series.

The nth term of a geometric series is given by ar^(n-1). Here, a = 4, r = 2, n = 6. So, 4 * 2^(6-1) = 4 * 32 = 128.

The sum of the first n terms of a geometric series is a(1 - r^n) / (1 - r). Here, a = 4, r = 2, n = 5. So, 4(1 - 2^5) / (1 - 2) = 4(1 - 32) / -1 = 124.

Substituting n = 7 into the formula gives a_7 = 3(7) + 2 = 21 + 2 = 23.

The sequence is 1, 2, 4, 8, 16, 32, 64. The 6th term is 64.

The sequence is 2, 3, 5, 8, 13. The 5th term is 13.

The sequence is 2, 3, 5, 8, 13. The fifth term is 13.

The first four terms are 1^2, 2^2, 3^2, 4^2 which are 1, 4, 9, 16. Their sum is 1 + 4 + 9 + 16 = 30.

The pattern is to add consecutive odd numbers: 2 + 3 = 5, 5 + 5 = 10, 10 + 7 = 17. The next term is 17 + 9 = 26.

For n = 4, a_4 = 4^2 + 2*4 = 16 + 8 = 24.

The common difference is found by subtracting any two consecutive terms: 15 - 10 = 5.

The sequence defined by a_n = 3n + 1 has a constant difference between consecutive terms, hence it is an arithmetic sequence.

The series consists of perfect squares: 1^2, 2^2, 3^2, 4^2, 5^2.